Math Help - Multinomial Coefficients: 2 Questions

1. Multinomial Coefficients: 2 Questions

1. Twenty-five girls are bored.. How many ways may the girls form two teams of 5 to play basketball against each other?

- This I used C(25,5) * C(20,5) and got 823,727,520. The answer in the back says it is 411,863,760. It might just be coincidence but the answer in the back is half the number I got.

2. A fraternity consisting of 30 members wants to play 7 vs. 7 flag football. How many different match-ups are possible?

-This I used C(30,7) * C(23,7) and got 4.99x10^11. The answer in the back says 2.495x10^11. Again, half my answer.

2. Re: Multinomial Coefficients: 2 Questions

Hello, kangta27!

I hope I can explain this clearly.

1. Twenty-five girls are bored. .How many ways may the girls
form two teams of 5 to play basketball against each other?

I used $C(25,5)\cdot C(20,5)$ and got $823,\!727,\!520.$
The answer in the back says it is 411,863,760.
It might just be coincidence but the answer in the back is half the number I got.

Let's name the twenty-five girls: Ann, Beth, Carol, Doris, . . . Yvonne.
Also, let's name the two teams: Reds and Blues.

$C(25,5) = 53,\!130$ is the number of ways to select the Red Team.

Then $C(20,5) = 15,\!504$ is the number of ways to select the Blue Team.

The product $(823,\!727,\!520)$ is the number of ways to form the two teams.

Included in this long list are these two partitions:

. . $\begin{array}{cc}\text{Red} & \text{Blue} \\ ABCDE & FGHIJ \end{array} \qquad\begin{array}{cc} \text{Red} & \text{Blue} \\ FGHIJ & ABCDE \end{array}$

Since the two teams do not have names,
. . these two cases represent the same division into two teams.

That is, our list has twice as many partitions than actually exist.

3. Re: Multinomial Coefficients: 2 Questions

Thanks Soroban! So in these situations, does that mean I always divide the answer by 2?

4. Re: Multinomial Coefficients: 2 Questions

Suppose you have a collection of n objects and you wish to form 2 groups from the set each with m objects where $2m\le n$.

By the commutativity of multiplication, it doesn't matter if you choose m items to go into the first group, then m items to go into the second group, or if you alternate in any manner. By the fundamental counting principle, we find the number of permutations is:

$n(n-1)(n-2)\cdots(n-(2m-2))(n-(2m-1))$

Now we must account for the permutations within the two groups themselves:

$\frac{n(n-1)(n-2)\cdots(n-(2m-2))(n-(2m-1))}{m!m!}=$

$\frac{n\cdots(n-(m-1))}{m!}\cdot\frac{(n-m)\cdots(n-(2m-1))}{m!}=$

$\frac{n!}{(n-m)!m!}\cdot\frac{(n-m)!}{(n-2m)!m!}=$

${n \choose m}\cdot{n-m \choose m}$

Now we must account for the permutations of the number of groups:

$\frac{1}{2!}{n \choose m}\cdot{n-m \choose m}=$

$\frac{1}{2}{n \choose m}\cdot{n-m \choose m}$

One could generalize to say that for p groups (where $pm\le n$) the number of ways N is:

$N=\frac{1}{p!}\prod_{k=0}^{p-1}{n-km \choose m}$