# GMAT Math vs. School Math – Part II - Counting

In my last article, we examined the importance of considering mathematical strategies other than those we learned in school. So, rather than approach a question as though our former teachers will be checking (and grading!!) every detail of our work to ensure the use of “appropriate” mathematical techniques, we should approach questions with the sole purpose of determining the correct answer as quickly as possible, regardless of whether or not those teachers would approve.

In this article, we’ll examine an often-maligned strategy that can be applied to all counting questions.

Listing and Counting

Perhaps it’s its simplicity that makes listing and counting so unpalatable for many students. Granted, on the surface, it seems like a strategy a 5-year-old might use, but the fact of the matter is that, if you’ve been staring at a counting question for over a minute, and you still have no idea where to begin, you should either make an educated guess or start listing possible outcomes. In many cases, the simple process of listing outcomes will help us gain valuable insight into a fast way to solve the question. With this in mind, consider the following 400-level counting question:

If repeated letters are allowed, how many two-letter words can we create using the 26 letters in the English language?

A) 52

B) (26)(25)

C) 26^2

D) 2^26

E) 26!

Let’s start listing possible outcomes in a systematic way.

Words beginning with A: AA, AB, AC, AD,...., AX, AY, AZ (26 in total)

Words beginning with B: BA, BB, BC, BD,...., BX, BY, BZ (26 in total)

Words beginning with C: CA, CB, CC, CD,...., CX, CY, CZ (26 in total)

At this point we should see a pattern.

Once we’ve assigned in the first letter, there are 26 possible words. Since there are 26 possible first letters, and 26 possible words for each first letter, the total number of words = 26 x 26 = 26^2 = C

Okay, now try the listing and counting approach with this 600-level question:

If repeated letters are allowed, how many 5-letter words can we create using only E’s and Q’s?

A) 10

B) 25

C) 32

D) 50

E) 120

Have you tried it yet? Did you have difficulty finding a systematic way to list all of the possible 5-letter outcomes? If so, we might find it easier to start with 1-letter words, then 2-letter words, and so on.

1-letter words: E, Q (2 in total)

2-letter words: EE, EQ, QE, QQ (4 in total)

3-letter words: EEE, EEQ, EQE, EQQ, QEE, QEQ, QQE, QQQ (8 in total)

Aha! Each time we add a letter, the number of possible words doubles.

This means that there will be 16 4-letter words, and 32 5-letter words.

So, the correct answer is C.

This last one is a 750-level question (at least). See how you do.

How many positive integers less than 1000 are such that the sum of their digits is 6?

A) 18

B) 28

C) 30

D) 36

E) 48

In order to list the possible outcome in a systematic way, let’s treat all of the possible outcomes as 3-digit numbers where the hundreds digit and the tens digit can equal 0. So, for example, we’ll write 15 as 015, and we’ll write 6 as 006.

With this in mind, here’s one way to list the possible outcomes:

Integers beginning with 0: 006, 015, 024, 033, 042, 051, 060 (7 in total)

Integers beginning with 1: 105, 114, 123, 132, 141, 150 (6 in total)

Integers beginning with 2: 204, 213, 222, 231, 240 (5 in total)

See the pattern yet?

Integers beginning with 3: 303, 312, 321, 330 (4 in total)

So, the total number of outcomes = 7+6+5+4+3+2+1 = 28 = B

Aside: For a different (counting-based) approach to the above question, check out this post: http://www.beatthegmat.com/very-tricky-counting-problem-t25349.html

The question is slightly different, but the approach is the same.

Final Words

As you might imagine, listing and counting won’t be the best (i.e., fastest) strategy for every counting question. What’s important, however, is that you understand its potential value and consider it a worthy candidate when considering your options.