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A class with 25 students, 15 women and 10 men.
A committe will be formed by 3 students, a president, a vice president and public relation manager. How many mixed committee can be formed?
So, what I did, there are 15 women and 10 men, so 15*10 ,and there will be left 23 students, then there are 3! places for them so 15*10*23*3!

I think that this would work however it's wrong, another solution is(doing by cases):
$$ 15*(^10A_2)*3 + 10*(^15A_2)*3 $$
Notation: $$^nA_p=^nP_p $$
They're the number of arrangements/permutations.

Simplifying their answer we get:
$$15*10*9*3 + 10*15*14*3 = 15*10*3(9+14)=15*10*23*3$$
Well the difference is, they have 3 but I have 3!, what's wrong with my answer? Why do I have to divide by two?

So, what I did, there are 15 women and 10 men, so 15*10, and there will be left 23 students ...

In this step you made a mistake:

What you do here is forming the committee by doing three steps:

1. You pick one of 15 women (let's say Jane Doe)


2. You pick one of 10 men (let's say John Doe)


3. You pick one of the 23 remaining students (let's say Joe Bloggs)

However, you could also have picked Joe Bloggs in the second step and John Doe in the third step.

This means that you will get each possible set of three persons twice when you pick the three persons this way.

Therefore there are only $frac12*15*10*23$ sets of three persons, not 15*10*23.

Well the difference is, they have 3 but I have 3!

When the persons in the committee have different tasks, it is correct to multiply the number of sets of persons with 3! (and not with 3) to get the number of possible different committees.

This will lead to the following result:

$(frac12*15*10*23)*(3!) = frac12*15*10*23*(3!) = 15*10*23*frac12*6 = 15*10*23*3$

Say men are $M_1,M_2,...,M_10$ and women are $W_1,W_2,...W_15$. Then in your method, if we choose a woman first, say $W_1$, then a man, say $M_1$, then from the rest of $23$, suppose we have chosen $M_2$. And then we permute them with $3!$ so it is not important in which order we have chosen this three member. But notice that if we choose $W_1$ as before as woman and $M_2$ as man this time, and $M_1$ from the remaining $23$ students, permuting them will give us the same committee, resulting in overcounting.

In general, if we choose $W_i$ and $M_j$ first and then $W_k$, this committee can be formed in exactly $2$ ways in your method because we can either choose $W_i$ first and $W_k$ third, or $W_k$ first $W_i$ third. This is also valid for $M_k$ too so it is actually valid for all committees formed in your method. Therefore, we are counting each case exactly twice. That's why your answer is twice the correct answer.

Upon clarifying that we are speaking of a triple, then these can be 1M 2W or 2M 1W.

Then they can be permuted each in $3!$ ways, thus a total of $12$ if identity does not matter.

On the contrary, If people identity matters, in the first case you can select the members in $binom101 cdot binom152$, in the second in $binom102 cdot binom151$, and permute the roles.

i.e. $3! (binom101 cdot binom152+binom102 cdot binom151)= 10350$

Suppose you select a woman, call her A, and a man, call him B, then the third person, call them C. When you multiply by 3!, you form every possible permutation of A,B,C.

Now suppose you select, without loss of generality, C, then B, then A. When you multiply this triplet by 3!, you are forming the same collection of committees.

Since you are double counting, you need to divide by 2.