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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.