# 2010 HSC Question (1 Viewer)

#### leehuan

##### Well-Known Member
4d

A group of 12 people is to be divided into discussion groups.

(i) In how many ways can the discussion groups be formed if there are 8 people in one group, and 4 people in another?

(ii) In how many ways can the discussion groups be formed if there are 3 groups containing 4 people each?

So the answer to part (i) is \binom{12}{8} which is kinda trivial
The answer to part (ii) is \binom{12}{4}*\binom{8}{4} / 3!. I eventually convinced myself that the division by 3! was because of repetition/double counting from the ordering being unimportant.

But then I had a look at part (i) again and I got confused. Suddenly I'm not sure why division by 1/2! was not necessary for part (i) whereas division by 1/3! was necessary for part (ii). Clarification please?

#### InteGrand

##### Well-Known Member
4d

A group of 12 people is to be divided into discussion groups.

(i) In how many ways can the discussion groups be formed if there are 8 people in one group, and 4 people in another?

(ii) In how many ways can the discussion groups be formed if there are 3 groups containing 4 people each?

So the answer to part (i) is \binom{12}{8} which is kinda trivial
The answer to part (ii) is \binom{12}{4}*\binom{8}{4} / 3!. I eventually convinced myself that the division by 3! was because of repetition/double counting from the ordering being unimportant.

But then I had a look at part (i) again and I got confused. Suddenly I'm not sure why division by 1/2! was not necessary for part (i) whereas division by 1/3! was necessary for part (ii). Clarification please?
In part (ii), the groups are indistinguishable, so we over-count by a factor of 3! if we just do (12C4)*(8C4). For part (i), the two groups are indistinguishable, so there's no over-counting.

To see why there's overcounting in part (i), note that what we're doing with (12C4)*(8C4)*(4C4) is effectively choosing 4 people for group A, then 4 people for group B, then 4 people for group C, where groups A,B,C each have four people. So by doing (12C4)*(8C4)*(4C4), we're counting something like these as different groupings:

- Group A = {1,2,3,4}, Group B = {5,6,7,8}, Group C = {9,10,11,12}
- Group A = {5,6,7,8}, Group B = {1,2,3,4}, Group C = {9,10,11,12}
(where the people are called by numbers 1 to 12).

As we can see, the above are actually the same groupings since we don't actually have any distinction between Groups A,B,C. So for any partitioning into groups of four people, we over-count by 3! if we do (12C4)*(8C4)*(4C4) (since for each partitioning, we can arrange them into "A,B,C" in 3! ways). Hence the answer is found by dividing (12C4)*(8C4)*(4C4) by 3!.

#### leehuan

##### Well-Known Member
In part (ii), the groups are indistinguishable, so we over-count by a factor of 3! if we just do (12C4)*(8C4). For part (i), the two groups are indistinguishable, so there's no over-counting.

To see why there's overcounting in part (i), note that what we're doing with (12C4)*(8C4)*(4C4) is effectively choosing 4 people for group A, then 4 people for group B, then 4 people for group C, where groups A,B,C each have four people. So by doing (12C4)*(8C4)*(4C4), we're counting something like these as different groupings:

- Group A = {1,2,3,4}, Group B = {5,6,7,8}, Group C = {9,10,11,12}
- Group A = {5,6,7,8}, Group B = {1,2,3,4}, Group C = {9,10,11,12}
(where the people are called by numbers 1 to 12).

As we can see, the above are actually the same groupings since we don't actually have any distinction between Groups A,B,C. So for any partitioning into groups of four people, we over-count by 3! if we do (12C4)*(8C4)*(4C4) (since for each partitioning, we can arrange them into "A,B,C" in 3! ways). Hence the answer is found by dividing (12C4)*(8C4)*(4C4) by 3!.
Ah ok hmm.

In that case would there have been overcounting if we still had two groups (for 12 people), but wanted 6 people in each group then?

#### InteGrand

##### Well-Known Member
Ah ok hmm.

In that case would there have been overcounting if we still had two groups (for 12 people), but wanted 6 people in each group then?
$\bg_white \noindent Yep. In general if we have N = ng people and want to split them up into g groups of n people each, the no. of ways is: \binom{N}{n}\times \binom{N-n}{n}\times \binom{N-2n}{n}\times \cdots \times \binom{2n}{n}\times \binom{n}{n}\bigg/ g!. I.e. if dividing into \emph{equally sized groups}, then we need to \emph{divide by the factorial of the no. of groups}, for the same reason as above (over-counting since we can permute the groups in g! ways).$

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