arranging p numbered balls in n indistinguishable boxes

Question

let pDn the number of ways to arrange p numbered balls in n indistinguishable boxes such that p is greater than or = n and no box is empty how to calculate pDp-1. (I tried to dissect the problem and I think it's (pDn=(p)(p-1)/2!) but I can't prove it formaly, and I'm not sure about my answer...help please)

Answer 1

When calculating pDp-1 consider that there is exactly one more balls than boxes. Thus, if no box is empty, there is exaclty one box with two balls, and the rest have one exactly.

Since the boxes are indistinguishable, to count the number of ways to arrange the balls, we need only consider how many ways there are to pair two balls together.

The ways of choosing 2 balls from p is $$\frac{p!}{2!(p-2)!}$$

This can be written in the simpler form $$\frac{p(p-1)}{2}$$