Posted on
24
Mar 2021

Standard Deviation Problem On The GMAT (Normative Distribution)

Standard Deviation 700+ Tips

Hey guys! Today, we’re going to take a look at a standard deviation problem. And standard deviation is a concept that only comes up infrequently on the GMAT. So it’s more important to have a basic understanding of the concepts associated with it than to go really deep. This is true largely for much of the statistics, probability, and combinatorics problems. They show up infrequently until you get to the higher levels and even when you’re at the higher levels, relative to the algebra, arithmetic and even the geometry problems, they play such a small role.

And yet there’s so much math there that it’s very easy to get caught up in spending a lot of time prepping on problems or on these types of math that offer very little in return relative to spending your prep time really mastering the things that come up frequently. I’m not saying don’t learn this stuff I’m saying balance it according to its proportionality on the GMAT. As a general rule you can assume that stat, combinatorics and probability, all that advanced math, constitutes maybe 10 to 15% of what you’ll see on the GMAT. So keep that in mind as you prep.

Problem Language

In this problem the first step is to figure out what the heck we’re actually being asked for and it’s not entirely clear. This one’s written a little bit in math speak. So we have a normal distribution which doesn’t really matter for this problem but if you studied statistics it just means the typical distribution with a mean m in the middle and a standard deviation of d which they tell us is a single standard deviation. So they’re really just telling us one standard deviation but they’re saying it in a very tricky way and they’re using a letter d. If it helps you can represent this graphically.

Graphical Representation

Notice that they tell you something that you may already know: that one standard deviation to either side of the mean is 68 in a normal distribution. This breaks up to 34, 34. But they’re asking for everything below. The +1 side of the distribution. Since the m, the mean is the halfway point, we need to count the entire lower half and the 34 points that are in between the mean and the +d, the +1 standard deviation. This brings us to 84 which is answer choice D. This is primarily a skills problem, that is, you just need to know how this stuff works. There’s no hidden solution path and the differentiation done by the GMAT here is based upon your familiarity with the concept. Rather than heavy-duty creative lifting as we see on so many other problems that have more familiar math, that everyone kind of knows.

I hope this was helpful. Check out below for other stat and cool problem links and we’ll see you guys next time. If you enjoyed this GMAT Standard Deviation problem, try this Data Sufficiency problem next.

Posted on
23
Mar 2021

Consecutive Integers (plus more on Odds and Evens)

By: Rich Zwelling, Apex GMAT Instructor
Date: 23rd March, 2021

In our last post, we left you with a GMAT Official Guide Data Sufficiency problem to tackle regarding Odd/Even Number Theory. Here it is, if you didn’t get a chance to do it before:

If x and y are integers, is xy even?
(1) x = y + 1.
(2) x/y is an even integer.

Solution:

The title of today’s post gives a little hint. We discussed last time that for a product of integers to be even, all you need is a single even integer in the set. (Conversely, for the entire product to be odd, every integer must be odd.)

How does that affect how we interpret Statement (1)? Well, this is where taking a purely Algebraic approach can get you into trouble. Why not take a NARRATIVE APPROACH? What is the equation really telling us narratively about the relationship between x and y? Develop the  habit of thinking about numbers narratively instead of purely algebraically, as this can make numerical relationships easier to understand.

Statement (1), in essence, is really telling us that y and x are consecutive integers. If I take y and add one to it to get x, they must be consecutive. So what are the implications for the number property of the question (even/odd)? Well, between any pair of consecutive integers, one must be odd and one must be even. I don’t know which is which, but it doesn’t matter in this case, because I care only about the product. Whether it’s Odd*Even or Even*Odd, the final product of xy will always be Even.

Even if you don’t see this narratively, you could use a scenario-driven approach and test simple numbers using the equation. If x = y + 1, try y = 2 and x = 3 to get xy = 6. Now, that’s just one instance of xy being even, so that doesn’t prove it’s always even. But then if I use y = 3 and x = 4 to get xy = 12, I again get an even result for xy. Using y = 4 and x = 5 would again yield an even xy result. Hopefully what I will realize, at this point, is that I am switching between x=odd, y=even and x=even, y=odd, and yet I still end up with xy=even

Statement (1) is SUFFICIENT.

And remember: the GMAT is very fond of testing interesting properties of consecutive integers.

For Statement (2), we discussed that division is less amenable to hard-fast rules of odd/even properties. For that reason, you could definitely use a scenario-driven approach. But hopefully, this approach would lead you towards a consideration of our previously discussed undesired possibility. Here’s what I mean:

If x/y is even, does that guarantee that xy is even? If you’re picking numbers, you must pick ones that fit the statement. It’s tempting to pick ones that contradict the statement (i.e. proving the statement wrong), but remember that the question is up for debate, not the statement. The statement is given to you as iron-clad fact.

So what if, for example, x = 4 and y = 2? That works for Statement (2), because x/y would certainly be even (4/2 = 2). And that would lead you to xy = 8, which is of course even.

You could pick many such examples. But here’s where the undesired part comes into play. Is it possible for us to pick numbers here such that xy becomes odd? Well, for xy to be odd, both x and y would have to be odd. But if x and y were both odd, could x/y be even as Statement (2) says? Even if you can’t see the answer right away, try some numbers here, knowing that x and y must be integers:

x=5, y=35/3 (not even)
x=3, y=5 → 3/5  (not even)
x=3, y=7 → 3/7 (not even)
x=9, y=3 → 3 (not even)

You’ll see that no matter what numbers you try, you’ll never get an even result for x/y. From a number theory perspective, this is because to get an even result, you must retain a factor of 2 in the numerator of the fraction. But we don’t even start with a factor of 2 if we have only odd numbers to begin with.

In conclusion, there’s no way that x and y can both be odd in Statement (2), meaning xy is not odd, and that guarantees xy is even. Statement (2) is also SUFFICIENT.

The correct answer is D. Each statement is SUFFICIENT on its own.

For next time, give the following Official Guide problem a shot, and use it as a chance to practice a NARRATIVE APPROACH:

The sum of 4 different odd integers is 64. What is the value of the greatest of these integers?
(1) The integers are consecutive odd numbers
(2) Of these integers, the greatest is 6 more than the least.

Posted on
18
Mar 2021

Odds and Ends (…or Evens)

By: Rich Zwelling, Apex GMAT Instructor
Date: 18th March, 2021

Last time, we signed off with an Official Guide GMAT problem that provided a nice segue into Number Theory, specifically today’s topic of GMAT Odds and Evens. Now we’ll discuss the solution. Here’s the problem, in case you missed it and want to try it now:

If x is to be chosen at random from the set {1, 2, 3, 4} and y is to be chosen at random from the set {5, 6, 7}, what is the probability that xy will be even?

A) 1/6
B) 1/3
C) 1/2
D) 2/3
E) 5/6

Method #1 (Certainly passable, but not preferable)

Since there are so few numbers involved here, you could certainly take a brute-force approach if pressed for time and unsure of a faster strategy. It doesn’t take long to map out each individual product of x*y systematically and then tally up which ones are even. Here I’ll use red to indicate not even and green to indicate even:

1*5 = 5
1*6 = 6
1*7 = 7

2*5 = 10
2*6 = 12
2*7 = 14

3*5 = 15
3*6 = 18
3*7 = 21

4*5 = 20
4*6 = 24
4*7 = 28

Since we know that all probability is Desired Outcomes / Total Possible Outcomes, and since we have 8 even results out of 12 total possible outcomes, our final answer would be 8/12 or 2/3.

However, this is an opportune time to introduce something about odd and even number properties and combine it with the method from our “undesired” probability post…

Method #2 (Far preferable)

First, some number theory to help explain:

You might have seen that there are rules governing how even and odd numbers behave when added, subtracted, or multiplied (they get a little weirder with division). They are as follows for addition and subtraction:

Even ± Odd = Odd

Odd ± Odd = Even

And with multiplication, the operative thing is that, when multiplying integers, just a single even number will make the entire product even. So for example, the following is true:

Odd * Odd * Odd * Odd * Odd * Odd * Odd * Odd * Odd * … * Odd = Odd

But introduce just a single even number into the above product, and the entire product becomes even:

Even * Odd * Odd * Odd * Odd * Odd * Odd * Odd * Odd * Odd * … * Odd = Even

This makes sense when you think about it, because you are introducing a factor of 2 to the product, so the entire product must be even

When considering the above rules, you could memorize them, or you could turn to SCENARIO examples with simple numbers to illustrate the general pattern. For example, if you forget what Odd * Odd is, just multiply 3 * 5 to get 15, which is odd, and that will help you remember.

This can also help you see that introducing just a single even number makes an entire product even:

3 * 3 * 3 = 27

2 * 3 * 3 * 3 = 54

So how does all of that help us get the answer to this problem faster?

Well, the question asks for the desired outcome of xy being even. That presents us with three possibilities:

1. x could be even AND y could be odd
2. x could be odd AND y could be even
3. x could be even AND y could be even

This is why I, personally, find it less helpful to think of individual multiplication rules for even numbers and more helpful to think in terms of: “The only way a product of integers is odd is if every integer in the set is odd.”

Because now, we can just think about our undesired outcome, the only outcome left:

4. x is odd AND y is odd, making the product xy odd

And how many such outcomes are there in this problem? Well at this point, we can treat it like a simple combinatorics problem. There are two odd numbers in the x group (1 and 3) and two odd numbers in the y group (5 and 7):

_2_ * _2_ = 4 possible odd xy products

And for the total, there are four possible x values and three possible y values:

_4_ * _3_ = 12 total possible xy products

That gives us a 4/12 or 1/3 probability of getting our undesired odd xy product. And as discussed in the previous post, we can now simply subtract that 1/3 from 1 to get:

1 – 1/3  = 2/3 probability that we get our desired even xy product.

For a little “homework,” try the following Official Guide GMAT problem. It has an underlying topic that we’ll discuss next time:

If x and y are integers, is xy even?
(1) x = y + 1.
(2) x/y is an even integer.

If you enjoyed this GMAT odds and evens article watch Mike solve this Number Theory problem with multiple solution paths.

Posted on
16
Mar 2021

When Probability Meets Combinatorics: One Problem, Two Approaches

By: Rich Zwelling, Apex GMAT Instructor
Date: 16th March, 2021

Now, we’d like to take a look at an Official GMAT Probability problem to pull everything together. The following is a good example for two reasons:

1. It illustrates a quirky case that is difficult more conceptually than mathematically, and thus is better for the GMAT.

2. It can be tackled either through straight probability or through a combination of probability and combinatorics.

Here’s the question:

Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter, she prepared an envelope with its correct address. If the 4 letters are to be put into the 4 envelopes at random, what is the probability that only 1 letter will be put into the envelope with its correct address?

A) 1/24
B)
1/8
C) 1/4
D) 1/3
E) 3/8

First, as always, give the problem a shot before reading on for the explanation. If possible, see if you can tackle it with both methods (pure probability and probability w/ combinatorics).

Explanation #1:

First, we’ll tackle pure probability. Let’s label the letters A, B, C, and D, and let’s say that A is the letter we wish to match with its correct envelope. The other three will be matched with incorrect envelopes. We now must examine the individual probabilities of the following events happening (green for correct, red for incorrect):

_A_   _B_   _C_   _D_

For the above, each slot represents a letter matched with an envelope. There are four envelopes and only one is correct for letter A. That means Tanya has a 1/4 chance of placing letter A in its correct envelope:

_1/4__   _B_   _C_   _D_

We now desire letter B to be placed in an incorrect envelope. Two of the remaining three envelopes display incorrect addresses, so there is a 2/3 chance of that happening:

_1/4__   _2/3_   _C_   _D_

We then desire letter C to also be placed in an incorrect envelope. Only one of the remaining two envelopes displays an incorrect address, so there is a 1/2 chance of that happening:

_1/4__   _2/3_   _1/2_   _D_

At that point, the only remaining option is to place the last remaining letter in the last remaining envelope (i.e. a 100% chance, so we place a 1 in the final slot):

_1/4__   _2/3_   _1/2_   _1_

Multiplying the fractions, we can hopefully see that some cancelling will occur:

¼ x ⅔ x ½ x 1

= 1 x 2 x 1
———–
4 x 3 x 2

= 1/12

But lo and behold, 1/12 is not in our answer choices. Did you figure out why?

We can’t treat letter A as the only possible correct letter. Any of the four letters could possibly be the correct one. However, the good news is that in any of the four cases, the math will be exactly the same. So all we have to do is take the original 1/12 we just calculated and multiply it by 4 to get the final answer: 4 x 1/12 = 4/12 = 1/3. The correct answer is D.

Explanation #2:

So what about a combinatorics approach?

As we’ve discussed in our previous GMAT probability posts, all probability can be boiled down to Desired Outcomes / Total Possible Outcomes. And as we discussed in our posts on GMAT combinatorics, we can use factorials to figure out the total possible outcomes in a situation such as this, which is actually a simple PERMUTATION. There are four envelopes, so for the denominator of our fraction (total possible outcomes), we can create a slot for each envelope and place a number representing the letters in each slot to get:

_4_  _3_  _2_  _1_  =  4! = 24  possible outcomes

This lets us know that if we were to put the four letters into the four envelopes at random, as the problem says, there would be 24 permutations, giving us the denominator of our fraction (total possible outcomes).

So what about the desired outcomes? How many of those 24 involve exactly one correctly placed letter? Well, let’s again treat letter A as the correctly placed letter. Once it’s placed, there are three slots (envelopes) left:

___  ___  ___

But the catch is: the next envelope has only two letters that could go into it. Remember, one of the letters correctly matches the envelope in address, and we want a mismatch:

_2_  ___  ___

Likewise, that would leave two letters available for the next envelope, but only one of them would have the wrong address:

_2_  _1_  ___

And finally, there would be only one choice left for the final envelope:

_2_  _1_  _1_

That would mean for the correctly-placed A letter, there are only two permutations in which each of the other letters is placed incorrectly:

_2_ x  _1_ x  _1_ = 2 possible outcomes.

But as before, we must consider that any of the four letters could be the correct letter, not just letter A. So we must multiply the 2 possible outcomes by four to get 8 desired outcomes involving exactly one letter being placed in its correct envelope. That gives us our numerator of Desired Outcomes. Our denominator, remember, was 24 total possible outcomes. So our final answer, once again, is 8/24 = 1/3.

This is a great example of how GMAT combinatorics can intersect with probability.

To tide you over until next time, give this Official GMAT problem a try. It will also give a nice segue into Number Theory, which we’ll begin to talk more about going forward. Explanation next time…

If x is to be chosen at random from the set {1, 2, 3, 4} and y is to be chosen at random from the set {5, 6, 7}, what is the probability that xy will be even?

A) 1/6
B) 1/3
C) 1/2
D) 2/3
E) 5/6

Posted on
11
Mar 2021

An “Undesired” Approach to GMAT Probability

By: Rich Zwelling, Apex GMAT Instructor
Date: 11th March, 2021

In our last post, we discussed a solution for the following question, which is a twist on an Official Guide GMAT probability problem:

Xavier, Yvonne, and Zelda individual probabilities for success on a certain problem are 1/4, 1/2 and 5/8, respectively. Xavier will attempt the problem first. If he solves it, Yvonne and Zelda will not attempt it. If Xavier cannot solve it, Yvonne will attempt it next. If she solves it, Zelda will not attempt it. If Yvonne cannot solve it, Zelda will then attempt it. What is the probability that Zelda does not get to attempt the problem?

A) 3/16
B)
5/8
C) 3/8
D) 5/64
E) 3/64

We also mentioned that there was an alternate way to solve it. Did you find it? In truth, it relates to something we discussed in a previous post we did on GMAT Combinatorics, specifically Combinations with Restrictions. In that post, we discussed the idea of considering combinations in which you’re not interested. It might seem counterintuitive, but if you subtract those out from the total number of combinations possible, you’re left with the number of combinations in which you are interested:

You can actually do something similar with probability. Take the following basic example:

Suppose I told you to flip a fair coin five times, “fair” meaning that it has an equal chance of landing heads-up or tails-up. I then wanted to know the probability that I flip at least one head. Now, when you think about it, the language “at least one” involves so many desired possibilities here. It could be 1 head, 2 heads, …, all the way up to 5 heads. I’d have to calculate each of those probabilities individually and add them up.

Or…

I could consider what is not desired, since the possibilities are so much fewer:

All of the above must add to 100% or 1, meaning all possible outcomes. So why not figure out the probability that I get 0 heads (or all tails), and then subtract it from 100% or 1 (depending on whether I’m using a percentage or decimal/fraction)? I’ll then be left with all the possibilities in which I’m actually interested, without the need to do any more calculations.

Each time I flip the coin, there is a ½ chance that I flip a tail. This is the same each of the five times I flip the coin. I then multiply all of the probabilities together:

½ x ½ x ½ x ½ x ½ = 1 / 25  = 1 / 32

Another way to view this is through combinatorics. Remember, probability is always Desired outcomes / Total possible outcomes. If we start with the denominator, there are two outcomes each time we flip the coin. That means for five flips, we have 25 or 32 possible outcomes, as illustrated here with our slot method:

_2_  _2_  _2_  _2_  _2_ = 32

Out of those 32 outcomes, how many involve our (not) desired outcome of all tails? Well, there’s only one possible way to do that:

_T_  _T_  _T_  _T_  _T_    ← Only 1 outcome possible

It really is that straightforward: with one outcome possible out of 32 total, the probability is 1/32 that we flip all tails.

Now remember, that is our, not desired. Our desired is the probability of getting at least one head

So, since the probability of getting 0 heads (all tails) is 1/32, we simply need to subtract that from 1 (or 32/32) to get our final result. The probability that we flip at least one head if we flip a fair coin five times is 31/32.

Application to problem from previous post

So now, how do we work that into the problem we did last time? Well, in the previous post, we took a more straightforward approach in which we considered the outcomes we desired. But can we use the above example and consider not desired instead? Think about it and give it a shot before reading the explanation:

Xavier, Yvonne, and Zelda individual probabilities for success on a certain problem are 1/4, 1/2 and 5/8, respectively. Xavier will attempt the problem first. If he solves it, Yvonne and Zelda will not attempt it. If Xavier cannot solve it, Yvonne will attempt it next. If she solves it, Zelda will not attempt it. If Yvonne cannot solve it, Zelda will then attempt it. What is the probability that Zelda does not get to attempt the problem?

A) 3/16
B) 5/8
C) 3/8
D) 5/64
E) 3/64

Explanation

In this question, our desired case is that Zelda does not attempt the problem. That means, quite simply, that our not desired case is that Zelda does get to attempt it. This requires us analytically to consider how such a case would arise. Let’s map out the possibilities with probabilities:

Notice that two complementary probabilities are presented for each box. For example, since there is a 1/4 chance Xavier solves the problem (left arrow), we include the 3/4 probability that he does not solve the problem (right arrow).

If Zelda does get to attempt it, it’s clear from the above that first Xavier and Yvonne must each not solve it. There is a 3/4 and a 1/2 chance, respectively, of that happening. This is also a dependent situation. Xavier must not solve AND Yvonne must not solve. Therefore, we will multiply the two probabilities together to get ¾ x ½ = ⅜. So there is a 3/8 chance of getting our not desired outcome of Zelda attempting the problem.

So, we can finally subtract this number from 1 (or 8/8) and see that there is a 5/8 chance of Zelda not getting to attempt the problem. The correct answer is B.

Next time, we’ll discuss how GMAT Probability and Combinatorics can combine to form some interesting problems…

Posted on
09
Mar 2021

Independent vs. Dependent Probability

By: Rich Zwelling, Apex GMAT Instructor
Date: 8th March, 2021

Independent vs. Dependent Probability

As promised last time, we’ll return to some strict GMAT probability today. Specifically, we’ll discuss the difference between independent and dependent probability. This simply refers to whether or not the events involved are dependent on one another. For example, let’s take a look at the following Official Guide problem:

Xavier, Yvonne, and Zelda each try independently to solve a problem. If their individual probabilities for success are 1/4, 1/2 and 5/8, respectively, what is the probability that Xavier and Yvonne, but not Zelda, will solve the problem?

A) 11/8
B)
7/8
C) 9/64
D) 5/64
E) 3/64

In this case, we are dealing with independent events, because none of the probabilities affect the others. In other words, what Xavier does doesn’t affect Yvonne’s chances. We can treat each of the given probabilities as they are.

So mathematically, we would multiply, the probabilities involved. (Incidentally, the word “and” is often a good indication that multiplication is involved. We want Xavier AND Yvonne AND not Zelda to solve the problem.) And if Zelda has a chance of solving the problem, that means she has a chance of not solving it.

The answer would therefore be ¼ x ½ x ⅜  = 3/64 or answer choice E.

What if, however, we changed the problem to look like this:

Xavier, Yvonne, and Zelda individual probabilities for success on a certain problem are 1/4, 1/2 and 5/8, respectively. Xavier will attempt the problem first. If he solves it, Yvonne and Zelda will not attempt it. If Xavier cannot solve it, Yvonne will attempt it next. If she solves it, Zelda will not attempt it. If Yvonne cannot solve it, Zelda will then attempt it. What is the probability that Zelda does not get to attempt the problem?

A) 3/16
B)
5/8
C) 3/8
D) 5/64
E) 3/64

As you can see, the problem got much more complicated much more quickly, because now, the question stem is dependent upon a very specific series of events. Now, the events affect one another. Xavier will attempt the problem, but what happens at this stage affects what happens next. If he solves it, everything stops. But if he doesn’t, the problem moves to Yvonne. So in effect, there’s a ¼ chance that he’s the only person to attempt the problem, and there’s a ¾ chance the problem moves to Yvonne.

This is most likely how the GMAT will force you to think about probability: not in terms of formulas or complicated mathematical concepts, but rather in terms of narrative within a new problem with straightforward numbers.

That brings us to consideration of the question stem itself. What would have to happen for Zelda not to attempt the problem? Well, there are a couple of possibilities:

1. Xavier solves the problem

If Xavier solves the problem, the sequence ends, and Zelda does not see the problem. This is one case we’re interested in, and there’s a ¼ chance of that happening.

2. Xavier does not solve, but then Yvonne solves

There’s a ½ chance of Yvonne solving, but her seeing the problem is dependent upon the ¾ chance that Xavier does not solve. So in reality, we must multiply the two numbers together to acknowledge that the situation we want is “Xavier does not solve AND Yvonne does solve.” This results in ¾ x ½ = ⅜

The two above cases constitute two independent situations that we now must add together. For Zelda not to see the problem, either Xavier must solve it OR Yvonne must solve it. (The word “or” is often a good indication that addition will be used).

This leads us to our final probability of ¼ + ⅜ = that Zelda does not get to attempt the problem.

There is an alternative way to solve this problem, which we’ll talk about next time. It will segue nicely into the next topic, which we’ve already hinted at in our posts on GMAT combinatorics. Until then…

Posted on
04
Mar 2021

Combinations with Restrictions

By: Rich Zwelling, Apex GMAT Instructor
Date: 4th March, 2021

In our previous post, we discussed how GMAT combinatorics problems can involve subtracting out restrictions. However, we discussed only PERMUTATIONS and not COMBINATIONS.

Today, we’ll take a look at how the same technique can be applied to COMBINATION problems. This may be a bit more complicated, as you’ll have to use the formula for combinations, but the approach will be the same.

Let’s start with a basic example. Suppose I were to give you the following problem:

The board of a large oil company is tasked with selecting a committee of three people to head a certain project for the following year. It has a list of ten applicants to choose from. How many potential committees are possible?

This is a straightforward combination problem. (And we know it’s a COMBINATION situation because we do not care about the order in which the three people appear. Even if we shift the order, the same three people will still comprise the same committee.)

We would simply use the combination math discussed in our Intro to Combination Math post:

10!
10C3 =       ————-
3! (10-3!)

10!
———
3! (7!)

10*9*8
———
3!

10*9*8
———
3*2*1

= 120 Combinations

However, what if we shifted the problem slightly to look like the following? (As always, give the problem a shot before reading on…):

The board of a large oil company is tasked with selecting a committee of three people to head a certain project for the following year. It has a list of ten applicants to choose from, three of whom are women and the remainder of whom are men. How many potential committees are possible if the committee must contain at least one woman?

A) 60
B) 75
C) 85
D) 90
E) 95

In this case, there’s a very important SIGNAL. The language “at least one” is a huge giveaway. This means there could be 1 woman, 2 women, or 3 women which means we would have to examine three separate cases. That’s a lot of busy work.

But as we discussed in the previous post, why not instead look at what we don’t want and subtract it from the total? In this case, that would be the case of 0 women. Then, we could subtract that from the total number of combinations without restrictions. This would leave behind the cases we do want (i.e. all the cases involving at least one woman).

We already discussed what happens without restrictions: There are 10 people to choose from, and we’re selecting a subgroup of 3 people, leading to 10C3  or 120 combinations possible.

But how do we consider the combinations we don’t want? Well, we want to eliminate every combination that involves 0 women. In other words, we want to eliminate every possible committee of three people that involves all men. So how do we find that?

Well, there are seven men to choose from, and since we are choosing a subgroup of 3, we can simply use 7C3 to find the number of committees involving all men:

7!
7C3 =       ————-
3! (7-3!)

7!
———
3! (4!)

7*6*5
———
3!

= 7*5 = 35 Combinations involving all men

So, out of the 120 committees available, 35 of them involve all men. That means 120-35 = 85 involve at least one woman. The correct answer is C.

Next time, we’ll return to probability and talk about how the principle of subtracting out elements that we don’t want can aid us on certain questions. Then we’ll dovetail the two and talk about how probability and combinatorics can show up simultaneously on certain questions.

Posted on
02
Mar 2021

Permutations with Restrictions

By: Rich Zwelling, Apex GMAT Instructor
Date: 2nd March, 2021

So far, we’ve covered the basics of GMAT combinatorics, the difference between permutations and combinations, some basic permutation and combination math, and permutations with repeat elements. Now, we’ll see what happens when permutation problems involve conceptual restrictions that can obscure how to approach the math.

To illustrate this directly, let’s take a look at the following Official Guide problem:

The letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT or DGIIT. Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?

A) 12
B) 18
C) 24
D) 36
E) 48

Did you catch the restriction? Up until the end, this is a standard permutation with repeats combinatorics problem, since there are five letters and two repeats of the letter ‘I’. However, we’re suddenly told that the two I’s must be separated by at least one other letter. Put differently, they are not allowed to be adjacent.

So how do we handle this? Well, in many cases, it’s helpful to set aside what we want and instead consider what we don’t want. It seems counterintuitive at first, but if we consider the number of ways in which the two I’s can appear together (i.e. what is not allowed) and then subtract that number from the total number of permutations without any restrictions, wouldn’t we then be left with the number of ways in which the two I’s would not appear together (i.e. what is allowed)?

Let’s demonstrate:

In this case, we’ll pretend this problem has no restrictions. In the word “DIGIT,” there are five letters and two I’s. Using the principle discussed in our Permutations with Restrictions post, this would produce 5! / 2! = 60 permutations.

However, we now want to subtract out the permutations that involve the two I’s side by side, since this condition is prohibited by the problem. This is where things become less about math and more about logic and conceptual understanding. Situationally, how would I outline every possible way the two I’s could be adjacent? Well, if I imagine the two I’s grouped together as one unit, there are four possible ways for this to happen:

II DGT

D II GT

DG II T

DGT II

For each one of these four situations, however, the three remaining letters can be arranged in 3*2*1 = 6 ways.

That produces a total of 6*4 = 24 permutations in which the two I’s appear side by side.

Subtract that from the original 60, and we have: 60 – 24 = 36. The correct answer is D

As you can see, this is not about a formula or rote memorization but instead about logic and analytical skills. This is why tougher combinatorics questions are more likely to involve restrictions.

Here’s another Official Guide example. As always, give it a shot before reading on:

Of the 3-digit integers greater than 700, how many have 2 digits that are equal to each other and the remaining digit different from the other 2 ?

(A) 90
(B) 82
(C) 80
(D) 45
(E) 36

Explanation

This is a classic example of a problem that will tie you up in knots if you try to brute force it. You could try writing up examples that fit the description, such as 717, 882, 939, or 772, trying to find some kind of pattern based on what does work. But as with the previous problem, what if we examine conceptually what doesn’t work?

This will be very akin to how we handle some GMAT probability questions. The situation desired is 2 digits equal and 1 different. What other situations are there (i.e. the ones not desired)?  Well, if you take a little time to think about it, there are only two other possibilities:

1. The digits are all the same
2. The digits are all different

If we can figure out the total number of permutations without restrictions and subtract out the number of permutations in the two situations just listed, we will have our answer.

First, let’s get the total number of permutations without restrictions. In this case, that’s just all the numbers from 701 up to 999. (Be careful of the language. Since it says “greater than 700”, we will not include 700.)

To get the total number of terms, we must subtract the two numbers then add one to account for the end point. So there are (999-701)+1 = 299 numbers in total without restrictions.

(Another way to see this is that the range between 701 and 999 is the same as the range between 001 and 299, since we simply subtracted 700 from each number, keeping the range identical. It’s much easier to see that there are 299 numbers in the latter case.)

Now for the restrictions. How many of these permutations involve all the digits being the same? Well, this is straightforward enough to brute force: there are only 3 cases, namely 777, 888, and 999.

How about all the digits being different? Here’s where we have to use our blank (or slot) method for each digit:

___ ___ ___

How many choices do we have for the first digit? The only choices we have are 7, 8, and 9. That’s three choices:

_3_  ___ ___

Once that first digit is in place, how many choices do we have left for the second slot? Well, there are 10 digits, but we have to remove the one already used in the first slot from consideration, as every digit must be different. That means we have nine left:

_3_  _9_  ___

Using the same logic, that leaves us eight for the final slot:

_3_  _9_  _8_

Multiplying them together, we have 3*9*8 = 216 permutations in which the digits are different.

So there are 216+3 = 219 restrictions, or permutations that we do not want. We can now subtract that from the total of 299 total permutations without restrictions to get our final answer of 299-219 = 80. The correct answer is C.

Next time, we’ll take a look at a few examples of combinatorics problems involving COMBINATIONS with restrictions.

Posted on
25
Feb 2021

What happens when Permutations have repeat elements?

By: Rich Zwelling, Apex GMAT Instructor
Date: 25th February, 2021

Permutations With Repeat Elements

As promised in the last post, today we’ll discuss what happens when we have a PERMUTATIONS situation with repeat elements. What does this mean exactly? Well, let’s return to the basic example in our intro post on GMAT combinatorics:

If we have five distinct paintings, and we want to know how many arrangements can be created from those five, we simply use the factorial to find the answer (i.e. 5! = 5*4*3*2*1 = 120). Let’s say those paintings were labeled A, B, C, D, and E.

Now, each re-arrangement of those five is a different PERMUTATION, because the order is different:

ABCDE

etc

Remember, there are 120 permutations because if we use the blank (or slot) method, we would have five choices for the first blank, and once that painting is in place, there would be four left for the second blank, etc…

_5_  _4_  _3_  _2_  _1_

…and we would multiply these results to get 5! or 120.

However, what if, say we suddenly changed the situation such that some of the paintings were identical? Let’s replace painting C with another B and E with another D:

ABBDD

Suddenly, the number of permutations decreases, because some paintings are no longer distinct. And believe it or not, there’s a formulaic way to handle the exact number of permutations. All you have to do is take the original factorial, and divide it by the factorials of each repeat. In this case, we have 5! for our original five elements, and we now must divide by 2! for the two B’s and another 2! for the two D’s:

5!
——
2! 2!

= 5*4*3*2*1
————-
(2*1)(2*1)

= 5*2*3
= 30 permutations

As another example, try to figure out how many permutations you can make out of the letters in the word BOOKKEEPER? Give it a shot before reading the next paragraph.

In the case of BOOKKEEPER, there are 10 letters total, so we start with a base of 10!

We then have two O’s, two K’s and three E’s for repeats, so our math will look like this:

10!
———
2! 2! 3!

Definitely don’t calculate this, though, as GMAT math stays simple and likes to come clean. Remember, we’ll have to divide out the repeats. You are extremely unlikely to have to do this calculation for a GMAT problem, however, since it relies heavily on busy-work mechanics. The correct answer choice would thus look like the term above.

Let’s now take a look at an Official Guide question in which this principle has practical use. I’ll leave it to you to discover how. As usual, give the problem a shot before reading on:

A couple decides to have 4 children. If they succeed in having 4 children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly 2 girls and 2 boys?

(A) 3/8
(B) 1/4
(C) 3/16
(D) 1/8
(E) 1/16

Quick Probability Review

Remember from our post of GMAT Probability that, no matter how complicated the problem, probability always boils down to the basic concept of:

Desired Outcomes
———————————–
Total Possible Outcomes

In this case, each child has two equally likely outcomes: boy and girl. And since there are four children, we can use are blank method to realize that we’ll be multiplying two 4 times:

_2_  _2_  _2_  _2_   =  16 total possible outcomes (denominator)

This may give you the premature notion that C or E must be correct, simply because you see a 16 in the denominator, but remember, fractions can reduce! We could have 4 in the numerator, giving us a fraction of 4/16, which would reduce to 1/4. And every denominator in the answer choices contains a factor of 16, so we can’t eliminate any answers based on this.

Now, for the Desired Outcomes component, we must figure out how many outcomes consist of exactly two boys and two girls. The trick here is to recognize that it could be in any order. You could have the two girls followed by the two boys, vice versa, or have them interspersed. Now, you could brute-force this and simply try writing out every possibility. However, you must be accurate, and there’s a chance you’ll forget some examples.

What if we instead write out an example as GGBB for two girls and two boys? Does this look familiar? Well, this should recall PERMUATIONS, as we are looking for every possible ordering in which the couple could have two girls and two boys. And yes, we have two G’s and two B’s as repeats. Here’s the perfect opportunity to put our principle into play:

We have four children, so we use 4! for our numerator, then we divide by 2! twice for each repeat:

4!
——
2! 2!

This math is much simpler, as the numerator is 24, while the denominator is 4. (Remember, memorize those factorials up to 6!)

This yields 6 desired outcomes of two boys and two girls.

With 6 desired outcomes of 16 total possible outcomes, our final probability fraction is 6/16, which reduces to 3/8. The correct answer is A.

Next time, we’ll look into combinatorics problems that involve restrictions, which can present interesting conceptual challenges.

Posted on
18
Feb 2021

A Continuation of Permutation Math

By: Rich Zwelling, Apex GMAT Instructor
Date: 16th February, 2021

Review of example from last post

Last time, when we started our discussion of GMAT Combinatorics, we gave a brief example of GMAT permutations in which we had five paintings and asked how many arrangements could be made on a wall with those paintings. As it turns out, no complicated combinatorics formula is necessary. You can create an easy graph with dashes and list five options for the first slot, leaving four for the second slot, and so on:

_5_  _4_ _3_ _2_ _1_

Then multiply 5*4*3*2*1 to get 120 arrangements of the five paintings. Remember you could see this notationally as 5!, or 5 factorial. (It’s helpful to memorize factorials up to 6!)

More permutation math

But there could be fewer slots then items. Take the following combinatorics practice problem:

At a cheese tasting, a chef is to present some of his best creations to the event’s head judge. Due to the event’s very bizarre restrictions, he must present exactly three or four cheeses. He has brought his best cheddar, brie, gouda, roquefort, gruyere, and camembert. How many potential orderings of cheeses can the chef create to present to the judge?

A) 120
B) 240
C) 360
D) 480
E) 600

First, as a review, how do we know this is a PERMUTATION and not a COMBINATION? Because order matters. In the previous problem, the word “arrangements” gave away that we care about the order in which items appear. In this problem, we’re told that we’re interested in the “orderings” of cheeses. Cheddar followed by gouda would be considered distinct from gouda followed by cheddar. (Look for signal words like “arrangements” or “orderings” to indicate a PERMUTATION problem.)

In this case, we must consider the options of three or four cheeses separately, as they are independent (i.e. they cannot both happen). But for each case, the process is actually no different from what we discussed last time. We can simply consider each case separately and create dashes (slots) for each option. In the first case (three cheeses), there are six options for the first slot, five for the second, and four for the third:

_6_  _5_  _4_

We multiply those together to give us 6*5*4 = 120 possible ways to present three cheeses. We do likewise for the four-cheese case:

_6_  _5_  _4_  _3_

We multiply those together to give us 6*5*4*3 = 360 possible ways to present four cheeses.

Since these two situations (three cheeses and four cheeses) are independent, we simply add them up to get a final answer of 120+360 = 480 possible orderings of cheeses, and the correct answer is D.

You might have also noticed that there’s a sneaky arithmetic shortcut. You’ll notice that you have to add 6*5*4 + 6*5*4*3. Instead of multiplying each case separately, you can factor out 6*5*4 from the sum, as follows:

6*5*4 + 6*5*4*3

= 6*5*4 ( 1 + 3)

= 6*5*4*4

= 30*16 OR 20*24

= 480

Develop the habit of looking for quick, efficient ways of doing basic arithmetic to bank time. It will pay off when you have to do more difficult questions in the latter part of the test.

Now that we have been through GMAT permutations, next time, I’ll give this problem a little twist and show you how to make it a COMBINATION problem. Until then…