In how many ways can we split $80$ persons in a $5$ wagon train such that : > In how many ways can we distribute $80$ persons in a $5$ wagon train such that : > > $a)$ exactly $15$ go into the first wagon > > $b)$ exactly $15$ go into one wagon For $a$) we have $\binom{80}{15}$ ways to choose $15$ persons for one wagon and for the rest of $65$ we use the stars and bars problem? How to approach $b)$ then ?

Assuming the people are distinguishable, as most people are...

(a) There are $\binom{80}{15}$ ways to choose the people which go in the first wagon, then $4^{65}$ ways to distribute the others (each person can go in one of four wagons).

(b) Use inclusion exclusion. For each wagon, count the number of ways where that particular wagon has $15$ people. Then subtract the doubly counted arrangements where two particular wagons have $15$ people, then add back in the triply counted ones, etc.