GMAT Math : Counting Methods

Study concepts, example questions & explanations for GMAT Math

Create An Account

All GMAT Math Resources

22 Diagnostic Tests 693 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

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:

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

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

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

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

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

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.

Report an Error

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:

22!

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

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

26!

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.

Report an Error

Example Question #1 : 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: $8 \times 5 = 40$

Report an Error

Example Question #3 : Counting Methods

How many subsets does a set with 12 elements have?

Possible Answers:

4096

132

$\small 12^{12}$

144

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

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

Report an Error

Example Question #1 : 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

427,518,000

71,253,000

121,500,000

729,000,000

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

Report an Error

Example Question #4 : Understanding Counting Methods

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

Possible Answers:

336

210

6,720

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:

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

Report an Error

Example Question #1 : 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 - 2,500

B= A + 2,500

B= A - 5,000

B= A + 5,000

B= A

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

Report an Error

Example Question #6 : 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 .

Report an Error

Example Question #2 : 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

5,940

40,320

11,880

495

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:

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

Report an Error

Example Question #1 : Understanding 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:

30,240

792

120

220

252

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:

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

Report an Error

All GMAT Math Resources

22 Diagnostic Tests 693 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Statistics Tutors in Houston, French Tutors in New York City, Math Tutors in Boston, Physics Tutors in Los Angeles, Math Tutors in Houston, Computer Science Tutors in Philadelphia, ACT Tutors in Denver, Biology Tutors in Houston, Calculus Tutors in Miami, Statistics Tutors in Seattle

Popular Courses & Classes

ISEE Courses & Classes in Miami, GMAT Courses & Classes in Boston, SAT Courses & Classes in Los Angeles, SSAT Courses & Classes in Washington DC, SAT Courses & Classes in Philadelphia, ACT Courses & Classes in Atlanta, SAT Courses & Classes in Boston, Spanish Courses & Classes in Phoenix, LSAT Courses & Classes in Los Angeles, GRE Courses & Classes in Denver

Popular Test Prep

GMAT Test Prep in Los Angeles, GRE Test Prep in Los Angeles, MCAT Test Prep in Washington DC, MCAT Test Prep in New York City, GMAT Test Prep in Boston, ISEE Test Prep in San Francisco-Bay Area, ACT Test Prep in Atlanta, LSAT Test Prep in Washington DC, GRE Test Prep in San Diego, SAT Test Prep in San Diego

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

Email address:
Your name:
Feedback:

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 #1 : Counting Methods

Example Question #3 : Counting Methods

Example Question #1 : Counting Methods

Example Question #4 : Understanding Counting Methods

Example Question #1 : Counting Methods

Example Question #6 : Understanding Counting Methods

Example Question #2 : Counting Methods

Example Question #1 : Understanding Counting Methods

All GMAT Math Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details