GMAT Math : Counting Methods

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Example Question #1 : Counting Methods

In how many ways can the 11th grade class elect a president, vice president, and treasurer from a class of 70 students?

Possible Answers:

$\dpi{100} \small 620,349$

$\dpi{100} \small 16,950$

$\dpi{100} \small 45,320$

$\dpi{100} \small 328,440$

$\dpi{100} \small 225,000$

Correct answer:

$\dpi{100} \small 328,440$

Explanation:

The president can be elected in 70 different ways. After a student is elected president, there are 69 students left to elect a vice president from. Similarly, there are then 68 students left for the spot of treasurer. So there are $\dpi{100} \small 70\times 69\times 68=328,440$ different arrangements.

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Example Question #1 : Understanding Counting Methods

What is the number of possible 4 letter code words that can be made from the alphabet, when all 4 letters must be different?

Possible Answers:

$\frac{26!}{22!}$

$\frac{26!}{4!}$

26!

22!

Correct answer:

$\frac{26!}{22!}$

Explanation:

This is a permutation of 26 objects (letters) taken 4 at a time. Here order matters, because for example, "abcd" is not the same code word as "bdca".

You must know the permutation formula! It is as follows:

$_{n}P_{r}=\frac{n!}{(n-r)!}$ , where n is the number of different objects taken r at a time.

Here we have $_{26}P_{4}=\frac{26!}{(26-4)!} = \frac{26!}{22!}$

Note: This is equivalent to 26 * 25 * 24 * 23.

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Example Question #2 : Counting Methods

There are 8 paths between places and and 5 paths betweeen places and . How many different routes are there between places and ?

Possible Answers:

170

336

Correct answer:

Explanation:

Multiple the number of routes for each piece of the trip: $8 \times 5 = 40$

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Example Question #4 : Understanding Counting Methods

How many subsets does a set with 12 elements have?

Possible Answers:

$\small 12^{12}$

4096

144

132

Correct answer:

4096

Explanation:

The number of subsets in a set of size is $\small 2^{n}$ . If n=12 , then the set has $\small \small \small \small 2^{12} = 4,096$ subsets.

Alternatively, each subset of this twelve-element set is essentially a sequence of 12 independent decisions, one per element - each decision has two possible outcomes, exclusion or inclusion. By the multiplication principle, this is 2 taken as a factor 12 times, or

$\small \small \small \small 2^{12} = 4096$

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Example Question #3 : Counting Methods

How many ways can a president, a vice-president, a secretary-treasurer, and three Student Senate representatives be selected from a class of thirty people? You may assume these will be six different people.

Possible Answers:

593,775

121,500,000

427,518,000

729,000,000

71,253,000

Correct answer:

71,253,000

Explanation:

This can be seen, without loss of generality, as choosing each officer in turn.

There are 30 ways of choosing the president; there are then 29 ways of choosing the vice-president, and 28 ways of choosing the secretary-treasurer. Then 3 Student Senate representatives are chosen from the remaining 27 students; this is a combination of 3 elements from 27 - that is, $C_{3}^{27}$ . By the multiplication principle, the number of possible selections of the officers is:

$N = 30 \cdot 29\cdot 28\cdot C_{3}^{27}$

$N= 30 \cdot 29\cdot 28\cdot \frac{ 27!}{(27-3)!\; 3!}$

$N= 30 \cdot 29\cdot 28\cdot \frac{ 27!}{24!3!}$

$N= \frac{30 \cdot 29\cdot 28\cdot 27 \cdot26\cdot25}{3\cdot 2\cdot 1}$

N= 71,253,000

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Example Question #6 : Understanding Counting Methods

How many ways can you select three different prime numbers between 1 and 20?

Possible Answers:

6,720

336

210

Correct answer:

Explanation:

There are eight prime numbers between 1 and 20:

$\left \{ 2,3,5,7,11,13,17,19\right \}$

The number of ways to select three of them, without regard to order, is the number of combinations of three out of eight:

$C(8,3) = \frac{8!}{3!(8-3)!} = \frac{8!}{3!5!} = \frac{6\cdot 7\cdot 8}{1\cdot 2\cdot 3}= 56$

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Example Question #7 : Understanding Counting Methods

A= 1-3+5-7+9-11+...+9,997-9,999

B = 2-4+6-8+10-12...+9,998 -10,000

Which of the following statements is true?

Possible Answers:

B= A

B= A + 5,000

B= A + 2,500

B= A - 2,500

B= A - 5,000

Correct answer:

B= A

Explanation:

The finite series is obtained from by increasing each term by 1; since is an alternating series, this results in adding to :

(1-1) + (1-1) + (1-1) + ...(1-1) = 0

so B = A + 0 = A

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Example Question #8 : Understanding Counting Methods

Define set $A = \left \{ 1,2,3,4,5,6,7,8\right \}$ .

How many four-element subsets of include at least three even numbers?

Possible Answers:

Correct answer:

Explanation:

Only one subset of has four even numbers - that is the subset $\left \{ 2,4,6,8\right \}$ .

Forming a subset with three even numbers and one odd number can be restated as choosing which even number to leave out and which odd number to include. There are 4 choices each way, so the number of sets fitting this description is $4\times 4 = 16$ .

Therefore, the number of subsets with at least three even elements is 16 + 1 = 17 .

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Example Question #4 : Counting Methods

Twelve students are running for student council; each student will vote for four. Mick wants to vote for his sister Janine. How many ways can he cast his ballot so as to include Janine among his choices?

Possible Answers:

165

40,320

495

11,880

5,940

Correct answer:

165

Explanation:

Since one of Mick's choices is already decided, he will choose three people from a set of eleven without regard to order. This is a combination of three from a set of eleven; the number of such combinations is:

$C (11,3) = \frac{11!}{(11-3)!3!}= \frac{11!}{8!3!} = \frac{11 \times 10 \times 9}{3 \times 2\times 1} = \frac{990}{6} = 165$

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Example Question #5 : Counting Methods

Twelve students are running for student council; each student will vote for five. Claude does not want to vote for Gary or Mitch, neither of whom he likes. How many ways can Claude fill in the ballot so that he does not vote for Gary?

Possible Answers:

252

30,240

220

120

792

Correct answer:

252

Explanation:

Claude will choose five people from a set of ten - twelve minus the two he dislikes - without regard to order. This is a combination of five from a set of ten; the number of such combinations is:

$C (10,5) = \frac{10!}{(10-5)!\; 5!} = \frac{10!}{ 5!\; 5!} = \frac{3,628,800 }{ 120 \times 120} = 252$

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GMAT Math : Counting Methods

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #1 : Counting Methods

Example Question #1 : Understanding Counting Methods

Example Question #2 : Counting Methods

Example Question #4 : Understanding Counting Methods

Example Question #3 : Counting Methods

Example Question #6 : Understanding Counting Methods

Example Question #7 : Understanding Counting Methods

Example Question #8 : Understanding Counting Methods

Example Question #4 : Counting Methods

Example Question #5 : Counting Methods

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