A school teacher wants to divide the whole class into 3 groups with...

Question

A school teacher wants to divide the whole class into 3 groups with an equal number of students in each group. The total number of students in her class is 24. Find out in how many ways can this be done?

A.  ^{24}C_8

B.  \frac{^{24}C_8 * ^{16}C_8 * ^8C_8}{3!}

C.  \frac{^{24}C_8 * ^{16}C_8 * ^8C_8}{3}

D.  ^{24}C_8 * ^{16}C_8 * ^8C_8

E.  ^{24}C_8 * ^{16}C_8 * ^8C_8  * 3!

Thanks,
Saquib
Quant Expert
e-GMAT

To practise ten 700+ Level Number Properties Questions attempt the The E-GMAT Number Properties Knockout .Click here.

vitaliyGMAT · Answer

\frac{24!}{(8!)^3 * 3!} = \frac{_{24}C_8*_{16}C_8*_8C_8}{3!}

Answer B

vitaliyGMAT · Answer

\frac{24!}{(8!)^3 * 3!} = \frac{_{24}C_8*_{16}C_8*_8C_8}{3!}

Answer B

That seems to be an identical question

chetan2u · Answer

Hi

The most important thing here :- all groups are identical

In 3 equal groups means 24/3 that is 8 in each group..
Choose 8 from total 24C8..
Choose 8 from remaining 16C8..
And finally remaining 8 are choosen 8C8

Total ways = 24C8 * 16C8 * 8C8..
But since the groups are identical ans these three groups can be arranged in 3!, Divide the ways by 3!

Ans B

Nitish7 · Answer

chetan2u

Hi Chetan2u,

We are referring the groups identical because the no. of students are same?

chetan2u · Answer

YES NITISH..

since the groups have same number and there is no other difference, these groups are identical..

example ..
there are three bags- same colour and same size- all bags are identical..
but say the bags are different in colour. not identical.. we will not divide by 31 or so

sasidharrs · Answer

Hi,
I understood till you came up to 24C8*16C8*8C8.. But, I don't know why are dividing by 3!.
We don't need to arrange the 3 groups right?, we just want to find how many ways to get 3 groups.
Lets take A,B,C are 3 groups.
We don't want to find ABC, ACB,CAB etc..

Clearly I am not following you. Can you please explain me

Thanks,

arpitkansal · Answer

chetan2u 
I don't understand why the groups are identical.The students are not identical so how can the groups be identical?
In such questions do we have to presume the groups as same identity and ignore the diversity of students?

chetan2u · Answer

Hi..
The reason is that we have not named the group..
So if A,B are in group I, C,D are in group 2 and E,F are in group 3..
It will be same when A,B are in group I, C,D are in group 3 and E,F are in group 2..
OR if A,B are in group 2, C,D are in group 1, and E,F are in group 3..

Since here all groups name and size wise are identical, all above distribution are identical, and hence we divide by 3!

Deebs · Answer

Correct if wrong:

The way I see it, there are two steps to this.

1) e.g. think about if we have 6 children divided into 3 groups. 2 kids in each.

How many kids in the first group? 6C2 - you have 6 kids and you choose 2 of them to be in the first group. 
How many kids in the second group? 4C2 - You already picked 2 kids to be in the first so 4 are remaining. choose 2 more out of 4. 
How many in the third? 2C2 - Only 2 kids left and choosing both of them.

So, we have 6C2*4C2*2C2. BUT...

2) Now, you have these groups (lets say, A,B,C). You can ofcourse arrange them in the order ABC. The question is, is the order CBA different from ABC? The answer is no. They all have 2 kids in them and even though the kids in them are different people, their combinations have already been accounted for in step 1. SO really, A=B=C. Ie. How many ways can you arrange AAA? 3*2*1/3!=1 right?

Same concept in this situation - 3 groups all identical and we found there are 6C2*4C2*2C2 ways of creating them so divide by 3! to get the total number of ways. Now you can apply the same to the problem above.

Thoughts?

DrAnkita91 · Answer

I answered as option E.

Please can anyone tell for these kind of questions... when should we divide by 3! and when should we multiply by 3!

chetan2u · Answer

When you have reptitions that have been counted multiple times then you require to remove them by dividing them by number that each combination is repeated.
Say, you have to select two groups consisting of equal members from A, B, C and D.
4C2*2C2 = 6 ways, but do we have 6 ways.
(1,2) = (AB,CD), (AC,BD), (AD,BC), 
So, where are the remaining 3 ways, they are reverse of the above, that is (CD,AB), (BD,AC) and (BC,AD).
And, why did this happen? Because we took 1 and 2 group to be different, but they are identical.
Thus divide by ways these groups can be arranged 2! for 2 groups and 3! for 3 groups or n! for n groups.

Now, when do you multiply by 3!?
arranging 3 out of 5 on three chairs.
You select three person from five in 5C3 ways, and arrange them in 3! ways within themselves, so multiply by 3! = 5C3*3!

DrAnkita91 · Answer

understood.. thanks

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

A school teacher wants to divide the whole class into 3 groups with...

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

A school teacher wants to divide the whole class into 3 groups with...

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards