From a group of 10 students, 7 girls and 3 boys, a teacher

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Gmat_mission: Legendary Member; Posts: 1622; Joined: Thu Mar 01, 2018 7:22 amFollowed by: 2 members

by Gmat_mission » Mon Jun 10, 2019 4:53 am

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From a group of 10 students, 7 girls and 3 boys, a teacher must choose 2 girls and 2 boys to present book reports. How many different arrangements of students, in order, are possible?

A. 252
B. 504
C. 1,008
D. 1,512
E. 5,040

[spoiler]OA=D[/spoiler]

Source: Princeton Review

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Source: Beat The GMAT — Problem Solving | Mon Jun 10, 2019 4:53 am

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Mon Jun 10, 2019 5:00 am

Gmat_mission wrote:From a group of 10 students, 7 girls and 3 boys, a teacher must choose 2 girls and 2 boys to present book reports. How many different arrangements of students, in order, are possible?

A. 252
B. 504
C. 1,008
D. 1,512
E. 5,040

Take the task of arranging students and break it into stages.

Stage 1: Select two girls
Since the order in which we select the women does not matter, we can use combinations.
We can select 2 girls from 7 girls in 11C2 ways (21 ways)
So, we can complete stage 1 in 21 ways

If anyone is interested, here's a video on calculating combinations (like 7C2) in your head: https://www.gmatprepnow.com/module/gmat- ... /video/789

Stage 2: Select two boys
We can select 2 boys from 3 boys in 3C2 ways (3 ways)
So, we can complete stage 2 in 3 ways

Stage 3: Arrange the 4 children in a row
We can arrange n unique objects in n! ways
So, we can arrange 4 unique children in 4! ways (=24 ways)
We can complete this stage in 24 ways.

By the Fundamental Counting Principle (FCP), we can complete all 3 stages (and thus select and arrange 4 children) in (21)(3)(24) ways (= 1512ways)

Answer: D
--------------------------

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmat- ... /video/775

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Brent Hanneson - Creator of GMATPrepNow.com

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8151; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

by Scott@TargetTestPrep » Tue Jun 11, 2019 7:09 pm

Gmat_mission wrote:From a group of 10 students, 7 girls and 3 boys, a teacher must choose 2 girls and 2 boys to present book reports. How many different arrangements of students, in order, are possible?

A. 252
B. 504
C. 1,008
D. 1,512
E. 5,040

[spoiler]OA=D[/spoiler]

Source: Princeton Review

We are selecting 2 of 3 boys. The number of boys can be selected in 3C2 = 3 ways.

We are selecting 2 of 7 girls. The number of girls can be selected in 7C2 = (7 x 6)/2! = 21 ways,

The total number of ways to select the group is 3 x 21 = 63.

The group of 4 children can then be ordered in 4! = 24 ways.

So, the total number of ways to select and then order the group is 63 x 24 = 1,512.

Alternate Solution:

First, let's find the number of ways to have the book reports presented by two girls, followed by two boys.

2 girls from a group of 7 girls can be selected and ordered in 7P2 = 7!/(7 - 2)! = 7 x 6 = 42 ways.

2 boys from a group of 3 boys can be selected and ordered in 3P2 = 3!/(3 - 2)! = 3 x 2 = 6 ways.

Thus, there are 42 x 6 = 252 ways to have two girls present their book report, followed by two boys.

Now, let's consider the different arrangements of genders to present the book report. We are looking for the arrangements of GGBB and by the indistinguishable permutations formula, there are 4!/(2!*2!) = (4 x 3)/2 = 6 of such arrangements.

Thus, the total number of different possible orderings for the presentation is 252 x 6 = 1,512.

Answer: D

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