arjtryarjtry wrote:

thanks bhushan.. it helped me.

so we have to divide b the no. of grps thus formed, and multiplication should be done for the same no. of grps is it ?

eg

three grps -> 6c2*4c2*2c2/3!

two grps-> 6c3*3c3/3!

suppose it was 1 group of 4 people

is it just 6c4?

Yes, if it is \(C_6^4\) for selecting 1 group of 4 people from 6. The thing that I forgot with regard to this question was the significance of "order does not matter." I was thinking that \(C_6^3\) was correct, but didn't remember to divide by the permutations of the order of the group. For instance.

3 groups of 3 from 9 people

\(\frac{C_3^9 * C_3^6 * C_3^3}{3!}\)

The first group formed can have any 3 of the 9. next group will be any 3 of the 6 remaining, and the last group is formed by selection the 3 from the 6, because you have 3 remaining, which is why it will always be 1. But then you have 3 groups, so you need to disregard their order, which means divide by the number of ways these 3 groups can be ordered. Which is 3!

Example:

A B C D E F G H I

CDE | BGI | AFH

This is 1 option. because order does not matter, we need to eliminate situations such as:

BGI | CDE | AFH

This is the same as counting the number of ways 3 things can be arranged. Permutation of 3 things = 3!.

I believe the answer for 3 groups of 3 out of 9 people, and order does not matter would be 280.

\(C_9^3 = \frac{9*8*7}{3*2*1} * C_6^3 = \frac{6*5*4}{3*2*1} * C_3^3 = 1\)

This all comes to be

\(\frac{9*8*7*6*5*4*1}{3*2*1*3*2*1*3*2*1}\)

\(8*7*5\)

\(7*40\)

\(280\)

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J Allen Morris

**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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