Total number of ways in which 256 identical balls can be placed in 16 numbered boxes ( 1 , 2 , 3 , . , 16 ) such that r th box contains at least r balls is ( 1 ≤ r ≤ 16 )

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The problem requires distributing 256 identical balls into 16 numbered boxes such that the $r^{t h}$ box contains at least $r$ balls, where $1 \leq r \leq 16$ .

First, we place the minimum required number of balls into each box. The first box gets 1 ball, the second box gets 2 balls, and so on, up to the 16th box which gets 16 balls. The total number of balls placed initially is:

$1 + 2 + 3 + \dots + 16 = r = 1 \sum 16 r = \frac{16 ( 16 + 1 )}{2} = \frac{16 \times 17}{2} = 8 \times 17 = 136$

We have placed 136 balls, so the remaining number of balls to be placed is $256 - 136 = 120$ . Now we need to distribute these 120 balls into the 16 boxes without any restrictions.

This is a stars and bars problem. We have 120 balls (stars) and 15 dividers (bars) to separate the balls into 16 boxes. The number of ways to arrange these is given by the combination formula:

$(k - 1 n + k - 1) = (n n + k - 1)$

where $n$ is the number of balls (120) and $k$ is the number of boxes (16).

So the number of ways to distribute the remaining 120 balls into 16 boxes is:

$(16 - 1 120 + 16 - 1) = (15 135)$

Therefore, the total number of ways to place the balls is $(15 135)$ .If we put minimum number of balls required in each box, then remaining 120 balls can be put in $^{135} C_{15}$ ways without any restriction.

Mathematics - Permutation & Combination Question with Solution | TestHub

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