lists with length $k$ with elements $b_1,b_2,\dots,b_k$ such that $|b_1|+|b_2|+···+|b_k| \le n$

Question

Ivan and Alexander write lists of integers. Ivan writes all the lists of length $n$ with elements $a_1,a_2,\dots,a_n$ such that $|a_1| + |a_2|+\dots+|a_n| \le k$. Alexander writes all the lists with length $k$ with elements $b_1,b_2,\dots,b_k$ such that $|b_1|+|b_2|+···+|b_k| \le n$. Prove that Alexander and Ivan wrote the same number of lists.

I was having trouble understanding the proof. I get how there are $\binom{k}{r}$ possible. But I don't get how there are $\binom{n}{r} 2^r$ ways, we have $r+1$ integers summing up to $k+1$. Also, I am not sure why total number of lists are $$\sum_{r=0}\binom{n}{r}\binom{k}{r}2^r.$$

We are introducing $c_0$ which is a positive element. So can someone explain the "bijective" way. As in, if we have elements $c_0,\dots,c_r$ summing to $k+1$. How do we get lists Ivan writes?

I think the author made a mistake as they are counting lists for Ivan and not Alexander.

Please help. Any new solution is also appreciated.

Answer 1

I'm going to assume that you do indeed, as you said, understand that there are $\binom{k}{r}$ ways to have $r$ positive integers with a sum $\le k$.

Remember that Alexander's list has $n$ elements, so the remaining $n-r$ elements in his list must all be zero's. Thus we have to distribute the $r$ positive integers among the $n$ elements in his list, and this can be done in $\binom{n}{r}$ ways.

Also, since the given restriction has absolute-values on each element, each of the $r$ positive integers could be replaced with its opposite, and this can be done in $2^r$ ways.

Thus the number of lists Alexander could write for a fixed value $r$ of non-zero entries is

$$\binom{k}{r} \binom{n}{r} 2^r$$

We have to sum this over all possible values of $r$; the minimum value for $r$ should be $1$, and $r$ cannot go over min$(k,n)$, so that would give us

$$\sum_{r=1}^{\min(k,n)} \binom{k}{r} \binom{n}{r} 2^r$$

If we use the same approach to calculate the number of lists Ivan can write, note the only difference would be the two binomials would be reversed, i.e.

$$\sum_{r=1}^{\min(k,n)} \binom{n}{r} \binom{k}{r} 2^r$$

so the total number of lists each can write is clearly the same.

In the given answer they extend the sum(s) to infinity by adopting the usual convention that $\binom{p}{q} = 0$ whenever $q>p$; I'm not sure why they did that, it's not wrong but it seems unnecessary. They're also starting the sums at $r=0$ which does seem wrong as there cannot be $0$ non-zero terms in either list, but since this simply adds $1$ to each of the two sums it still does not stop them from being equal.

Answer 2

Since in both cases there are $\leq$ conditions, we need to add a slack variable to each to convert them to equalities,

thus first list has $(k+1)$ elements totalling to $n$,
while the second has $(n+1)$ elements totalling to $k$

so by stars and bars, the first one has $\Large\binom{n+k}{k}$ number of lists, while the second has $\Large\binom{k+n}{n}$ which you should understand are identical numbers