From a group of three boys and four girls, a line of children arranged

Question

From a group of three boys and four girls, a line of children arranged from left to right is to be selected to salute the flag. If exactly five children are in the line, how many different such lines can be formed?

A. 28
B. 105
C. 252
D. 525
E. 2,520

A. 28
B. 105
C. 252
D. 525
E. 2,520

BrentGMATPrepNow · Answer

Notice that it doesn't make any difference that there are three boys and four girls to choose, since there are no restrictions regarding who can be in the 5-person line. 
So, let the 7 children be A, B, C, D, E, F, and G

Take the task of arranging 5 children and break it into   stages  .

Stage 1 : Select a child to be first in the line 
Since there are 7 children to choose from, we can complete stage 1 in  7  ways

Stage 2 : Select a child to be second in the line 
There are 6 remaining children from which to choose, so we can complete this stage in  6  ways.

Stage 3 : Select a child to be third in the line 
There are 5 remaining children from which to choose, so we can complete this stage in  5  ways.

Stage 4 : Select a child to be fourth in the line 
There are 4 remaining children from which to choose, so we can complete this stage in  4  ways.

Stage 5 : Select a child to be fifth in the line 
There are 3 remaining children from which to choose, so we can complete this stage in  3  ways.

By the Fundamental Counting Principle (FCP), we can complete all 5 stages (and thus create a 5-person line) in  (7)(6)(5)(4)(3)  ways (= 2520 ways)

Answer:  E

Note : the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn the technique.

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JeffTargetTestPrep · Answer

We are given that from a group of 3 boys and 4 girls, a line of 5 children arranged from left to right will be selected to salute the flag. We need to determine how many lines can be formed. Since we have an “order arrangement,” order matters. So we have a permutation problem. Thus, the number of ways to arrange 5 children from 7 is:

7P5 = 7!/(7-5)! = 7 x 6 x 5 x 4 x 3 = 2,520

Answer: E

EMPOWERgmatRichC · Answer

Hi All,

According to the prompt, there are no 'restrictions' in terms of who the five children in line will be, so the fact that the 7 total children are broken down into boys and girls is irrelevant. As such, this is just a straight-forward Permutation question and can be solved with basic multiplication.

For the 1st spot in line, there are 7 options... Once we place a child in the 1st spot....
For the 2nd spot in line, there are 6 options... Once we place a child in the 2nd spot....
For the 3rd spot in line, there are 5 options... Once we place a child in the 3rd spot....
For the 4th spot in line, there are 4 options... Once we place a child in the 4th spot....
For the 5th spot in line, there are 3 options... Once we place a child in the 5th spot....

(7)(6)(5)(4)(3) = 2520

You don't even have to do that full calculation though. Once you recognize that (4)(5) = 20, you know that you're looking for an answer that is a multiple of 20... and only one of the answers fits that pattern...

Final Answer:  E

GMATinsight · Answer

5 children need to be there
Case 1: 3 B and 2 G : 3C3*4C2*5! = 6*5!
Case 2: 2 B and 3 G : 3C2*4C3*5! = 12*5!
Case 3: 1 B and 4 G : 3C1*4C4*5! = 3*5!

Total ways to arrange = (6*5!) + (12*5!) + (3*5!) = (21*5!) = 21*120 = 2520

Answer: Option E

chetan2u · Answer

Hi

Do not waste time on looking at how many boys and girls are there as there are no RESTRICTIONS

So total are 3+4=7..
Choose 5 out of these 7.... 7C5
Arrange these 5 ... 5!

Total ways = 7C5*5! =  \frac{7!}{5!*2!}*5!=7*6*5*4*3 
Not reqd to calculate any further..
6*3 means product should be multiple of 9
5*4 means last two digits must be multiple of 4 and last digit 0..

ONLY E is left

Gmatcrusader007 · Answer

as it is an arrangement question we can find out the answer by

7P2 
=7!/2!
=7*6*5*4*3*2/2
=7*6*5*4*3
=2520

bumpbot · Answer

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From a group of three boys and four girls, a line of children arranged

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