greprep said:
Thank you so much. Would I use the same method for this problem:
"how many different committees of 7 people can be formed from a group of 10 people?"
No, that requires using a combination, or binomial coefficient. Let's look at how we can determine this number without simply using a formula.
We know that for the first position on the committee, we have 10 choices, for the second we have 9 choices and so forth down to 4 choices for the 7th and final position. Now, if the order in which the members of the committee mattered, the number of ways we could choose the committee would in fact use the fundamental counting principle. Let's call the number of ways to pick such a committee, where the order chosen matters, as $N_1$...we would have:
$$N_1=10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4=\frac{10!}{3!}=\frac{10!}{(10-7)!}$$
This is referred to as permutations, and if we choose $r$ objects from $n$, where $r\le n$, then we may use the special notation:
$$_rP_n=\frac{n!}{(n-r)!}$$
However, since the order does not matter in this case, we are only interested in finding how many ways 7 can be chosen from 10 (will call this number $N$), we need to divide the previous result by the number of ways to order 7 objects, which is $7!$...thus:
$$N=\frac{N_1}{7!}=\frac{10!}{7!(10-7)!}$$
For the general case, where we are choosing $r$ objects from $n$ objects, where $r\le n$, we would have:
$$N=\frac{n!}{r!(n-r)!}$$
In probability, this is a special formula, and is referred to as combinations, and uses the notations:
$$_nC_r={n \choose r}=\frac{n!}{r!(n-r)!}$$
So, the answer to the question you asked is:
$$N={10 \choose 7}=\frac{10!}{7!(10-7)!}=\,?$$
By the way, for future reference, please start a new thread for a new question...this way our threads don't potentially become convoluted and difficult to follow. :)
