If we have a piles of 10 pennies , a pile of 10 nickels, a pile of 10 dimes , a pile of 9 quarters and a pile of 8 half-dollar coins.
How many ways to pick 11 coins from those five piles of coins which must consists of all five different value of coins?
I figured out how to pick 11 coins from five piles of coins:
(5-1 +11) C (11) = 15 C 11 = 1365
We can approach this in the following manner:
Let $a$ be the number of pennies chosen,
$b$ be the number of nickels chosen,
$c$ be the number of dimes chosen,
$d$ be the number of quarters chosen,
$e$ be the number of half-dollar coins chosen.
According to the given condition,
$a+b+c+d+e=11$
But $a,b,c,d,e\geq 1$
So, substituting $p=a-1$, $q=b-1$, $r=c-1$, $s=d-1$, $t=e-1$,
We get ,
$p+q+r+s+t=6$ where $p,q,r,s,t\geq0$
The number of natural solutions will be ${(6+4) \choose 4}$ i.e. $\binom{10}{4} = 210$
We don't even have to worry about whether $a,b,c,d,e$ will exceed their limit of $10,10,10,9,8$ because we have chosen at least one from each pile which gives $max\{a,b,c,d,e\}$ possible $=7$.