Choose 3 people from a group of 10 My problem is the following : how many ways are there to form a commitee of $3$ people with a group of $10$ men ? My answer : We choose $3$ people; there are $10\times 9 \times 8=P(10,3)$ ways of doing that. Why is my reasoning wrong ? This obsiously should be $10\choose 3$ but I have no idea why.
 A: The reason for this is because if you have say A,B,C,D and you were to choose 2 then you can do this by selecting first one letter. For example A, with a there are 3 different combinations: AB, AC, AD. After A we can ignore A and move onto B since every letter has already been taken out with A. With B we have: BC, BD.
Lastly, since we are ignoring A and B there is just one last combination which is CD. Summing all of these together gives us 6 different combinations which is precisely nCr(4,2).
If you were selecting 3 different items from a list of 5 such as A,B,C,D,E you would follow a similar approach:
ABC Start with the first letter A and the second letter B
ABD Add all the combinations with B and D
ABE
ACD
ACE
ADE
BCD Ignore A and repeat process
BCE
BDE
CDE Ignore A and B and finish off
The reason behind the formula is because you have A nCr(number of letters,how many you are selecting-1) number of times times, B nCr(number of letters-1,how many you are selecting-1) and C nCr(number of letters-2,how many you're selecting -1).
A: What's wrong with $10 \times 9 \times 8$ is that you take the order of the people chosen into account.  Say you're counting $ABC$ and $ACB$ as different objects.  That's the number of permutations of picking three objects from ten objects.  To count the number of combinations $C_3^{10}$ instead of $P_3^{10}$, we have to divide the number of permutations $P_3^{10}$ by $P_3^3 = 3!$.
