What is the number of ways of expressing 4 as the sum of 5 nonnegative integers?

Question

I read a book and it says that this is equivalent to putting 4 balls in the 5 numbered baskets, thus the answer is 70 by the combination with repetitions. However I think it is ridiculously wrong because so many repetitions arise from this method. For example one ball in the first basket and 3 balls in the third basket is differently counted from three balls in the first basket and one ball in the third basket. However, both are just 0+0+0+1+3. So I am curious what the real answer is. Could anyone help me?

Answer 1

If you want to avoid repetitions, there are only 5 ways : $4=4$, $4=3+1$, $4=2+2$, $4=2+1+1$, $4=1+1+1+1$. And add zeros to these equalities to get a sum of 5 non-negative numbers.

If you allow repetitions, $4=4$ gives 5 solutions (5 positions for the 4), $4=3+1$ gives $2{5 \choose 2}$ solutions, $4=2+2$ gives $5 \choose 2$ possibilities, $4=1+1+1+1$ gives 5 possibilities and finally $4=1+1+2$ leads to $5 \times {4 \choose 2}$ choices. This gives finally $5+20+10+5+5\times6=70$ different possibilities.