We have to find the possible number of ways in which we can put twenty balls in five boxes so that the first box contains only one ball when repetition of distribution of balls is allowed.

We will use the concept of multiplication because there are twenty sub jobs dependent on each other and are performed one after the other.

The thing that is distributed is considered to have choices not the things to which we have to give them, it means that in this problem the balls have choices more precisely four choices are there for each ball and boxes won’t choose any because letters have the right to choose. But one box has the right to choose any one of the twenty balls, so the first box has twenty choices.

The number of ways in which we can put nineteen balls in four boxes where repetition of distribution is allowed

4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 = 4¹⁹

Hence the total number of ways in which we can put twenty balls in five boxes such that first box contains only one ball = 20 × 4¹⁹.

ILLUSTRATION –

To illustrate suppose there are five balls a, b, c, d, e to be put in three boxes A, B, C such that first box contains only one ball.

Let the first box be A and suppose we have combination of a, A , so the rest combinations would be 2 × 2 × 2 × 2 = 2⁴ , but we have more combinations like b, A ; c, A ; d, A ; e, A these combinations are five , so we add them at the end, and the final answer we get is = 5 × 2⁴.