Question

In how many ways can 2t + 1 identical balls can be placed in three distinct boxes so that any two boxes together contain more than the third.

Answer 1

Hint: In this question the concept of permutations and combinations will be used. We need to apply logic and general algebra to proceed through the problem. We will first find the total ways of arranging the balls and then subtract the ways where one box contains more balls than the sum of the other two. In order to select r unique things from a total number of n things, the number of ways are-
${}_{}^{\text{n}}{\text{C}}_{\text{r}}^{} = \dfrac{{{\text{n}}!}}{{\left( {{\text{n}} - {\text{r}}} \right)!{\text{r}}!}}$

Complete step-by-step answer:
Let the three boxes be A, B and C. We have to arrange the balls such that any two boxes together contain more than the third. So, first we will find the number of ways in which all the balls can be arranged in the three boxes, and then subtract the ways in which one box has more balls than the sum of the other two.

We know that the number of ways to distribute n things among r different things is given by-
${}_{}^{{\text{n}} + {\text{r}} - 1}{\text{C}}_{{\text{r}} - 1}^{}$
So the number of ways in which we can distribute the 2t - 1 ball in 3 boxes are-
${}_{}^{2{\text{t}} + 1 + 3 - 1}{\text{C}}_{3 - 1}^{} = {}_{}^{2{\text{t}} + 3}{\text{C}}_2^{}$...(1)
Now, we will subtract the cases in which one box has more balls than the sum of the other two. Let us assume that we put (t + 1) balls initially in box A. Now, the remaining t balls can be arranged in the three boxes without any constraints, because box A will always have more balls than the remaining two boxes.

So the number of ways in which we can distribute the t balls in 3 boxes are-
${}_{}^{{\text{t}} + 3 - 1}{\text{C}}_{3 - 1}^{} = {}_{}^{{\text{t}} + 2}{\text{C}}_2^{}$

This is repeated in boxes B and C as well. So the total ways are-
$3.{}_{}^{{\text{t}} + 2}{\text{C}}_2^{}$...(2)

When we subtract equations (1) and (2), we will get the number of ways in which balls can be placed in three distinct boxes so that any two boxes together contain more than the third. This is given by-
$= {}_{}^{2{\text{t}} + 3}{\text{C}}_2^{} - 3.{}_{}^{{\text{t}} + 2}{\text{C}}_2^{}$
$= \dfrac{{\left( {2{\text{t}} + 3} \right)\left( {2{\text{t}} + 2} \right)}}{2} - \dfrac{{3\left( {{\text{t}} + 2} \right)\left( {{\text{t}} + 1} \right)}}{2}$
$= \dfrac{{4{{\text{t}}^2} + 10{\text{t}} + 6 - 3{{\text{t}}^2} - 9{\text{t}} - 6}}{2}$
$= \dfrac{{{{\text{t}}^2} + {\text{t}}}}{2} = \dfrac{{{\text{t}}\left( {{\text{t}} + 1} \right)}}{2}$
This is the required answer.

Note: There is no direct method to solve this problem. We need to form the equations according to the question, and simplify it using general algebra. One common mistake is that students often forget the formula for combinations and permutations, which should be remembered. Also, it has been given that the balls are identical but the boxes are distinct. So, we need to worry about the different cases for the type of balls, but we need to divide each case for each of the three boxes, that is why we multiplied by 3 as shown in equation (2).