Question

The number of distinct ways of placing four indistinguishable balls into five distinguishable boxes is _____

Answer 1

Answer:

70

Step-by-step explanation:

Let a, b, c, d, e be the number of balls in each of the five (distinguishable) boxes.

We need the number of ways of assigning non-negative (0 or up) integer values to a, b, c, d, e such that a + b + c + d + e = 4, the total number of balls.  [Each assignment of balls corresponds to an assignment of values to a, b, c, d, e, and vice versa.]

We could try to list them:

0+0+0+0+4, 0+0+0+4+0, etc.

but we might make a mistake.

Better to use new variables A, B, C, D, E where

A = a+1, B = b+1, C = c+1, D = d+1, E = e+1

This way we get rid of the problem of "non-negative" since each of A, B, C, D, E is a positive (1 or up) integer.  The original problem is then equivalent to the number of ways of assigning positive integer values to these variables such that

A + B + C + D + E = 9   (5 more than before because we've added five 1's).

Imagine 9 1's lined up in a row.  There are 8 spaces between them.  We just have to choose 4 of those 8 spaces to put in a "+" sign.  For each such choice, we get a corresponding solution for A, B, C, D, E, and so a solution for a, b, c, d, e, and so an arrangement of balls in the boxes.

The number of ways of doing this is just

$\left(\begin{array}{c}8\\4\end{array}\right) = \frac{8\times 7\times 6\times 5}{4\times 3\times 2\times 1} = 70.$

(i.e. "8 choose 4", which apart from using the formula, could have been obtained from Pascal's Triangle.)