Determining probability of certain combinations Say I have a set of numbers 1,2,3,4,5,6,7,8,9,10 and I say 10 C 4 I know that equals 210. But lets say I want to know how often 3 appears in those combinations how do I determine that?
I now know the answer to this is $\binom{1}{1}$ $\binom{9}{3}$
I am trying to apply this to solve a problem in my math book. School is over so I cant ask my professor, Im just trying to get a head start on next year.
There are 4 numbers S,N,M,K
S stands for students in the class N the number of kids going on the trip M My friend circle including me K the number of my friends I need on the trip with me to enjoy myself.
I have to come up with a general solution to find out the probability I will enjoy the trip If I am chosen to go on it.
So far I came up with $\binom{S-1}{N-1}$ /($\binom{M-1}{K}$ $\binom{N-K-1}{S-K-1}$)
it works for cases like 
10 4 6 4 & 3 2 2 1 but dosent work for 10 10 5 3
any help is appreciated

Corrected:
First off, your fraction is upside-down: $\binom{S-1}{N-1}$ is the total number of groups of $N$ students that include you, so it should be the denominator of your probability, not the numerator. Your figure of $\binom{M-1}K\binom{N-K-1}{S-K-1}$ also has an inversion: it should be $\binom{M-1}K\binom{S-K-1}{N-K-1}$, where $\binom{S-K-1}{N-K-1}$ is the number of ways of choosing the $N-(K+1)$ students on the trip who are not you or your $K$ friends who are going.
After those corrections you have 
$$\frac{\binom{M-1}K\binom{S-K-1}{N-K-1}}{\binom{S-1}{N-1}}\;.$$
The denominator counts all possible groups of $N$ students that include you. The first factor in the numerator is the number of ways to choose $K$ of your friends, and the second factor is the number of ways to choose enough other people (besides you and the $K$ friends already chosen) to make up the total of $N$. However, this counts any group of $N$ students that includes you and more than $K$ of your friends more than once: it counts such a group once for each $K$-sized set of your friends that it contains.
To avoid this difficulty, replace the numerator by
$$\binom{M-1}K\binom{S-M}{N-K-1}\;;$$
now the second factor counts the ways to fill up the group with students who are not your friends, so the product is the number of groups of $N$ students that contain you and exactly $K$ of your friends. Of course now you have to add in similar terms for each possible number of friends greater than $K$, since you’ll be happy as long as you have at least $K$ friends with you: it need not be exactly $K$.
There are $\binom{M-1}{K+1}\binom{S-M}{N-K-2}$ groups of $N$ that include you and exactly $K+1$ of your friends, another $\binom{M-1}{K+2}\binom{S-M}{N-K-3}$ that contain you and exactly $K+2$ of your friends, and so on, and the numerator should be the sum of these terms:
$$\sum_i\binom{M-1}{K+i}\binom{S-M}{N-K-1-i}\;.\tag{1}$$
You’ll notice that I didn’t specify bounds for $i$. $\binom{n}k$ is by definition $0$ when $k>n$ or $k<0$, so we really don’t have to specify them: only the finitely many terms that make sense are non-zero anyway..
Alternatively, you can count the $N$-person groups that include you and fewer than $K$ of your friends and subtract that from the $\binom{S-1}{N-1}$ groups that include you; the different must be the number that include you and at least $K$ of your friends, i.e., the number that is counted by $(1)$. The number that include you and $i$ of your friends is $\binom{M-1}i\binom{S-M-i}{N-1-i}$, so the number of groups that include you and fewer than $K$ of your friends is
$$\sum_{i=0}^{K-1}\binom{M-1}i\binom{S-M}{N-1-i}\;.$$
This is going to be a shorter calculation than $(1)$ if $K$ is small.
