Question

Tom has $15$ ping-pong balls each uniquely numbered from $1$ to $15$ also has a red box, a blue box and a green box.
$\left(a\right)$ How many ways did Tom place the $15$ distinct balls into the three boxes so that no box is empty?
$\left(b\right)$ Suppose now that Tom has replaced $5$ ping-pong balls in each box. How many ways can he each $5$ balls from the three boxes so that he chooses at least one from each box?

Answer 1

(a) No. of ways$={}^{15}{C}_3\times 3!\times 3^{12}$

No. of ways to select three balls$\to {}^{15}{C}_3$

Placing the three balls such that one in each box $\to 3!$

Placing in other boxes$\to 3^{12}$

(b) Ways of collecting balls from the boxes: $\underset{1}{(3R,1B,1G)}$, $\underset{2}{(1R,3B,1G)}$, $\underset{3}{(1R,1B,3G)}$, $\underset{4}{(2R,2B,1G)}$, $\underset{5}{(2R,1B,2G)}$, $\underset{6}{(1R,2B,2G)}$

No. of ways$={}^5{C}_3\times {}^5{C}_1\times {}^5{C}_1\times 3(1,2,3)+{}^5{C}_2\times {}^5{C}_2\times {}^5{C}_1(4,5,6)$

$=750+1500=2250$.