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
From a group of 10 students, 7 girls and 3 boys, a teacher
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Take the task of arranging students and break it into stages.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
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
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We are selecting 2 of 3 boys. The number of boys can be selected in 3C2 = 3 ways.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 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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