Question

The number of ways to place 11 identical balls in three distinct boxes, so that anytwo boxes together will contain more balls than the other one​

Answer 1

Answer:

4, 4, 3

3, 3, 5

1, 5, 5

2, 4, 5

Answer 2

Answer:

There are $15$ different ways in which $11$ identical balls can be kept in $3$ distinct boxes such that any $2$ boxes together will contains more balls than the other one.

Step-by-step explanation:

The formula for finding out the number of ways in which $2t+1$ identical balls can be places in $3$ distinct boxes so that any $2$ boxes together will contains more balls than the third is : $\frac{t}{2} (t+1)$

In our questions, $2t+1=11$

$2t=10\\t=5$

Hence, the required number of ways $=\frac{5}{2} (6)$

$=15$

There are $15$ different ways in which $11$ identical balls can be kept in $3$ distinct boxes such that any $2$ boxes together will contain more balls than the other one.