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.\[\]
(a) How many ways Tom place the 15 distinct balls into the three boxes so that no box is empty?\[\]
(b) Suppose now that Tom has placed 5 ping-pong balls in each box. How many ways can he choose 5 balls from the three boxes so that he chooses at least one from each box?\[\]

Answer 1

Hint: We find the number of ways Tom can place 3 balls so that none of the boxes are empty by choosing 3 unique balls from 15 unique balls, then choose a box and put the rest 12 balls in all possible ways. We multiply to get the result for part (a). We see in part(b) that Tom can choose 2 balls each from 2 boxes and 1 ball from 1 box or he can choose 3 balls from 1 box and 1 ball from each from 2 boxes. We find a number of ways for both the cases and add them to get the result for part (b).

Complete step-by-step solution:
We are given that Tom has 15 ping-pong balls each uniquely numbered from 1 to 15. So the balls are not identical. He also has a red box, a blue box, and a green box which means the boxes are also not identical.\[\]
a) We see that Tom has to place 15 distinct balls into three distinct boxes so that no box is empty. Tom can select 3 unique balls from 15 unique balls in $^{15}{{C}_{3}}$ ways. Then Tom puts one ball in each box to ensure the none of the boxes are empty. He can do it in 3! Ways. He can then place the rest $15-3=12$ balls in ${{3}^{12}}$ ways. So we use the rule of product and conclude that Tom can place 15 distinct balls into the three distinct boxes in $^{15}{{C}_{3}}\times 3!\times {{3}^{12}}$ ways.\[\]
b)We are given that Tom has placed 5 ping-pong balls in each box. He has to choose 5 balls in a way where he has to choose at least one ball from each box. Tom can do choose 5 balls in following ways \[\]
Case-1: Tom can choose 2 balls from 2 boxes and 1 ball from one box. He can select the balls from the boxes in $^{5}{{C}_{2}}{{\times }^{5}}{{C}_{2}}{{\times }^{5}}{{C}_{1}}$ ways. There are three boxes Red (R), Blue(B) and green(G). He can choose 2 balls from red box and 2 balls from blue box and 1 ball from green box . We denote it as $\left( 2R,2B,1G \right)$. He can similarly place in the order $\left( 2R,2B,1G \right)$ and $\left( 2R,2B,1G \right)$. So there are 3 orders. We use rule of product and find the number of ways for case-1 as $3{{\times }^{5}}{{C}_{2}}{{\times }^{5}}{{C}_{2}}{{\times }^{5}}{{C}_{1}}$\[\]
Case-2: : Tom can choose 1 ball from 2 boxes and 3 ball from one box . He can select the balls from the boxes in $^{5}{{C}_{1}}{{\times }^{5}}{{C}_{1}}{{\times }^{5}}{{C}_{3}}$. He can do it in the following order $\left( 3R,1B,1G \right),\left( 1R,3B,1G \right),\left( 1R,1B,3G \right)$ . So there are 3 orders. We use rule of product and find the number of ways for case-2 as $3{{\times }^{5}}{{C}_{1}}{{\times }^{5}}{{C}_{1}}{{\times }^{5}}{{C}_{2}}$. \[\]
The required result is the sum of ways from case-1 and case-2. So the total numbers ways he can choose is

\[3{{\times }^{5}}{{C}_{2}}{{\times }^{5}}{{C}_{2}}{{\times }^{5}}{{C}_{1}}+3{{\times }^{5}}{{C}_{1}}{{\times }^{5}}{{C}_{1}}{{\times }^{5}}{{C}_{2}}=750+1500=2250\]

Note: We note that the balls are not identical. If they would have been the identical and the number of ways Tom can put them in three different boxes is equal to the number of positive integral solutions of $x+y+z=15$ which we can get using the formula $^{n+r-1}{{C}_{r-1}}$ where $n=15$ is the constant term and $r=3$ is the number of unknowns.