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 wrong ways Tom place the 15 distinct balls in 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: In this question, we will first find the number of ways by which Tom can place 3 balls so that none of the boxes are empty by choosing 3 unique balls from 15 unique balls. After that, we will choose a box and put the rest 12 boxes in all the possible ways. We will then multiply all the found values to get the answer for part (A). For part (B), 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 the remaining two boxes. We will find a number of ways for both cases and then add them together to get the result for part B.

Complete step by step answer:
Here we are given 15 ping pong balls each uniquely numbered from 1 to 15. We also have three boxes, red, blue, and green. So all balls and all boxes are nonidentical.
For part A.
Here we need to find a number of ways in which Tom can place 15 distinct balls into three distinct boxes so that no box is empty. Let us first fill three boxes with 1 ball each. So Tom can select 3 unique balls from 15 balls in a number of ways as $ {}^{15}{{C}_{3}} $ . (Number of ways in which r items can be selected from n items is given by $ {}^{n}{{C}_{r}} $ ).
Tom will put these balls in each box to ensure each box is non-empty. He can do it in 3! ways then. Now he can place these 12 balls in any way for 3 boxes. Hence total number of ways that he can put these 12 balls become $ \underset{\text{12 times}}{\mathop{3\times 3\times \cdots \cdots \times 3}}\,={{3}^{12}} $ .
Hence total ways become $ {}^{15}{{C}_{3}}\times 3!\times {{3}^{12}}=\dfrac{15!}{3!12!}\times 3!\times {{3}^{12}}=2730\times {{3}^{12}}\text{ ways} $ .
For part B.
Here we are given that Tom has placed 5 ping pong balls in each box. He has to choose 5 balls in ways where he has to choose at least one ball from each box. There will arise two cases for that.
Case I: He can choose 2 balls from 2 boxes and 1 ball from 1 box. Therefore he can select balls from boxes in number of ways: $ {}^{5}{{C}_{2}}\times {}^{5}{{C}_{2}}\times {}^{5}{{C}_{1}} $ . But the box are distinct too. So those 2 boxes can be any two and 1 box can be any one. Hence, we have 3 times the total ways. So for this case, our total ways become $ 3\times {}^{5}{{C}_{2}}\times {}^{5}{{C}_{2}}\times {}^{5}{{C}_{1}} $ .
Case II: He can choose 1 ball from 2 boxes and 3 balls from one box. Also, three boxes are there. So total ways become $ 3\times {}^{5}{{C}_{1}}\times {}^{5}{{C}_{1}}\times {}^{5}{{C}_{2}} $ .
Hence total number of ways becomes,
\[\begin{align}
  & 3\times {}^{5}{{C}_{2}}\times {}^{5}{{C}_{2}}\times {}^{5}{{C}_{1}}+3\times {}^{5}{{C}_{1}}\times {}^{5}{{C}_{1}}\times {}^{5}{{C}_{2}} \\
 & \Rightarrow 3\times {}^{5}{{C}_{2}}\times {}^{5}{{C}_{1}}\left( {}^{5}{{C}_{2}}+{}^{5}{{C}_{1}} \right) \\
 & \Rightarrow 3\times \dfrac{5!}{2!3!}\times \dfrac{5!}{4!1!}\left( \dfrac{5!}{2!3!}+\dfrac{5!}{4!1!} \right) \\
 & \Rightarrow 3\times 10\times 5\left( 10+5 \right) \\
 & \Rightarrow 150\left( 15 \right) \\
 & \Rightarrow 2250\text{ ways} \\
\end{align}\]

Note:
 Students should keep in mind all the possibilities before giving a final answer. For events occurring simultaneously, we use multiplication but for only one event out of some events will occur, we use addition. Note that $ {}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!} $ .