Ten telegenic contestants with a variety of personality

Thats really useful
thanks for the explanation.

I got C as well but could you explain what you mean by the additional statement you made?? And while I'm at it, A BIG THANK YOU for your contributions in this forum. It's amazing the amount of help you provide to anyone who is willing to look for it.

thanks for the question and thanks for the explanatin as well

The difference between the case when we have group #1 and #2 and the case when we don't, is explained for example here: in-how-many-different-ways-can-a-group-of-8-people-be-125985.html as well as in other links presented.

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Do not have any alternate explanation. Yours is perfect.

10C5 AND 5C5 --------> 10!/5! X 1 = 252

Hey Guys,

since I haven't had probabilty in high school, I don't get anything here. With the C and 5! and so on. Could anyone explain a little further! Would be great!

Hi All,

The last post in this thread (from unceldolan) points out that each of the explanations takes it for granted that the reader knows the Combination Formula. If that's the case, then the other explanations should suffice.

If you DON'T know the Combination Formula, then here's what it is and how to use it.

Any time a prompt asks for "groups" or "combinations" of things, then the order of the things DOES NOT MATTER.

For example, if a 2-person team consists of A and B, then A,B is the same as B,A --> thus, order does NOT matter. Mathematically though, you're allowed to count this team TWICE - A,B and B,A are the same team, so it should only be counted ONCE. The Combination Formula removes all of the "duplicates", leaving you with the unique combinations for whatever situation you're working with.

The Combination Formula itself is:

N!/[K!(N-K)!]

N = the total number of items/people
K = the size of the subgroup

Here, we have 10 people and we're asked to form 2 groups of 5.

For the first group, N = 10 and K = 5....

10!/[5!5!] = (10)(9)(8)(7)(6)/(5)(4)(3)(2)(1) = 256 unique groups of 5 people

Once you have formed that group of 5, then the remaining 5 form the other group (so there's no more math to do)

Final Answer:

GMAT assassins aren't born, they're made,
Rich

I got C as well. But l'm confused in one thing - why are we not diving 252 by 2 to avoid repetition here. Could someone please clarify?

Hi,

I would say the reason is that the tribes are named as A and B..
so say c,d,e,f,g are in A and h,i,j,k,l are in B..
this will be different to c,d,e,f,g are in B and h,i,j,k,l are in A

But as you correctly mentioned if it were just two groups with no difference, we would divide the numbers by 2..

Let's take the task of creating the teams and break it into stages.

Stage 1: Select two 5 contestants to be in tribe A
Since the order in which we select the contestants does not matter, we can use combinations.
We can select 5 contestants from 10 contestants in 10C5 ways
10C5 = (10)(9)(8)(7)(6)/(5)(4)(3)(2)(1) = 252
So, we can complete stage 1 in 252 ways

Stage 2: Place the remaining 5 people in tribe B
There's only 1 way to place all 5 remaining people in tribe B
So we can complete this stage in 1 way.

By the Fundamental Counting Principle (FCP), we can complete the 2 stages (and thus create 2 tribes of 5 contestants each) in (252)(1) ways (= 252 ways)

Answer: C

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

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The number of ways to select the first tribe is 10C5:

(10!)/[(10 - 5)! x 5!]

= (10 x 9 x 8 x 7 x 6)/(5 x 4 x 3 x 2 x 1)

= 3 x 2 x 7 x 6 = 252 ways

The next tribe can be selected in 5C5 = 1 way. So the total number of ways the two tribes can be selected is 252 x 1 = 252.

Answer: C

The number of ways to select the first tribe is 10C5:

10C5 = (10!)/(10 - 5)! x 5!

= (10 x 9 x 8 x 7 x 6)/(5 x 4 x 3 x 2 x 1)

= 3 x 2 x 7 x 6 = 252

The next tribe can be selected in 5C5 = 1 way.

So there are 252 x 1 = 252 ways to select the two tribes.

Answer: C

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