Problem.
Given a multiset $S=\lbrace \infty.a, \infty.b, \infty.c, \infty.d, \infty.e \rbrace$, where $a,b,c,d$ and $e$ are distinct. How many the 10-combinations from $S$ where $a$ and $c$ at least occured once, $b$ at least occured two times, and $d$ and $c$ not occured ?
Please give me opening hint. I'm confusing between it be 1 case, or divide in some cases because $c$ occured two times with different case in the problem. Thanks!
According to Wikipedia, in a combination the order of the selected things does not matter.
If we assume that the last $c$ in the description of the problem should be an $e$ then there are three admissible types $a$, $b$, $c$ left, and we should take $10$ of them. Take one $a$, two $b$s, and one $c$. Then we should take $6$ more of these three types in an arbitrary way. In how many ways can this be done? This is a stars and bars problem. The answer is ${8\choose2}$.