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

g,g,b,b can be arranged in 4c2 ways

(7*6*3*2*4c2)= 1512 option D

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. So, be sure to learn it.

RELATED VIDEOS

Hi Brent GMATPrepNow

I thought we just chose 2 boys out of 4 and 2 girls out of 7 and then we are done.

Why do we need stage 3? what is the logic behind it?

Question: How many different arrangements of students, in order, are possible?
The question specifies that order matters.

So, for example, if we have chosen Sue, Ann, Bob and Joe in stages 1 and 2, we still need to arrange these 4 people in order.
That is, the arrangement Sue-Ann-Bob-Joe is different from the arrangement Bob-Ann-Sue-Joe

Cheers,
Brent

Thanks Brent.

May I ask what if the questions DOES NOT state 'in order', what would be the case here?

Thanks

In that case, the wording might go something like this:

From a group of 10 students, 7 girls and 3 boys, a teacher must choose 2 girls and 2 boys to present book reports. In how many different ways can the teacher select 2 girls and 2 boys, if the order in which the children present does not matter?

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

Can anyone please explain how to understand that the solution is (7C2)*(3C2)*4!, but not (7P2)*(3P2).

The arrangement is of 2 boys and 2 girls. So once they are selected ordering of them presenting book report matters. The ordering does not matter during selection of boys and girls.
For eg out of three boys A,B and C. it does not matter if A,B or B,A are selected because in final book report presentation in both case you are going to end up having same kind of ordering.

Hope it clarifies your doubt.

Give kudos if you like the explanation.

Given:
Total number of students = 10 (7 girls & 3 boys)
The teacher must choose 2 girls and 2 boys to present book reports from these 10 students

To find:
The number of different arrangements of students - in order

Process:
- Number of ways of selecting 2 girls from 7 girls and 2 boys from 3 Boys= 7C2∗3C2= 63
- Now, the number of ways of arranging these 4 students = 4!
- The number of different arrangements of students, in order = 63 * 4! = 1512

Therefore, the correct answer is Option D

Hello guys I was wondering whether somebody could help me with this one,

so let's say we start from here 7*6*3*2 now we need to consider the fact that order matters so at this point I'm confused above many people have written GGBB so 4!/2!*2! will give us the correct result, BUT I was wondering why we are not multiplying by 4! , I mean GG and BB are 4 different people so shouldn't we consider all the aspects of order? What am I missing?

1st thing the teacher must do is CHOOSE:

Choose 2 Girls out of 7 Different Girls
AND
Choose 2 Boys out of 3 Different Girls

The teacher can do this in: 7 Choose 2 * 3 Choose 2 = Total No. of Different Combinations of Boys and Girls



AND then 2nd) she must Arrange Each 1 of these Groups of 4 Kids chosen

since they are all Different People, she can Arrange Each Group of 4 Kids in ---- 4! Ways


"7 Choose 2" * "3 Choose 2" * 4! = Count of Different Arrangements, in order, the teacher can have given the conditions


(7! / 5! * 2!) * (3) * (4!) =

7 * 6 / 2 * 3 * 24 =

21 * 3 * 24 =

63 * 24 = 1,512

Answer D

\(\)

to answer your question shortly, Yes, these are all Distinct People. Think of them as 3 Boys (A , B , and C) and 7 Girls (1 , 2 , 3 , 4 , 5 , 6 , and 7)

the way I like to thing about these problems is in 2 Stages:


1st) the teacher must CHOOSE from Each Distinct Group

Keep the 2 Groups Separate. When she chooses 2 Boys, she can only choose 2 Boys out of the 3 Available. When she Chooses 2 Girls, she can only choose 2 Girls out of the 7 Available.

Also keep in mind that when she's choosing, she must Choose 2 Boys AND Choose 2 Girls to complete the 4 kids.

So choose 2 Boys out of the 3

AND (and always means Multiply)

then choose 2 Girls out of the 7


2nd)For every different combination that you found above, the teacher is going to make a Different Arrangement. So if you multiply all the Combinations you found above by 4!, you will account for all the different arrangements (whether BBGG or GGBB or GBGB etc.)

I hope something was able to help....took me too long to wrap my head around these problems (and I'm still not there all the way)

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