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mleviko: Senior | Next Rank: 100 Posts; Posts: 63; Joined: Sun May 20, 2007 10:29 am; Thanked: 3 times

Combinations - from PR

by mleviko » Mon Feb 11, 2008 3:03 am

To fill a number of vacancies, an employer must hire 3 programmers from among 6 applicants and 2 managers from among 4 applicants.
What is the total number of ways in which she can make her selection?

A. 1,490
B. 132
C. 120
D. 60
E. 23

OA: C

Smashed the GMAT: 760

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Source: Beat The GMAT — Problem Solving | Mon Feb 11, 2008 3:03 am

smushkas: Senior | Next Rank: 100 Posts; Posts: 49; Joined: Sun Jul 01, 2007 5:56 pm; Thanked: 3 times

by smushkas » Mon Feb 11, 2008 3:25 am

This is a combination problem as you mentioned in the topic. Thus, we use the combination formula (order does not matter) to figure out what we need.
1) Programmers needed 3 out of 6 => 6! /[(6-3)! * 3!] = 120 / 6 = 20 ways
2) Managers needed 2 out of 4 => 4!/ [(4-2)! * 2!] = 12 / 2 = 6 ways

These 2 then multiplied, i.e. 20 * 6 = 120 ways of selection

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Jeff@TargetTestPrep: GMAT Instructor; Posts: 1462; Joined: Thu Apr 09, 2015 9:34 am; Location: New York, NY; Thanked: 39 times; Followed by:22 members

by Jeff@TargetTestPrep » Wed Dec 13, 2017 6:54 am

mleviko wrote:To fill a number of vacancies, an employer must hire 3 programmers from among 6 applicants and 2 managers from among 4 applicants. What is the total number of ways in which she can make her selection?

A. 1,490
B. 132
C. 120
D. 60
E. 23

This is a combination problem since the order of selecting the applicants for each job title does not matter.

3 programmers from 6 applicants can be hired in 6C3 ways:

6C3 = 6!/3!(6-3)! = (6 x 5 x 4)/3! = (6 x 5 x 4)/(3 x 2 x 1) = 20

2 managers from 4 applicants can be hired in 4C2 ways:

4C2 = 4!/2!(4-2)! = (4 x 3)/2! = (4 x 3)/(2 x 1) = 6

Thus, the total number of ways in which she can make her selection is 20 x 6 = 120.

Answer: C

Jeffrey Miller
Head of GMAT Instruction
[email protected]