Krdytkyu

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 $a,b,c,d,egeq 1$

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 $p,q,r,s,tgeq0$

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 $=7$.