Let q and ℓ be positive integers. Then the sum
\sum_{k=q}^\ell (-1)^{k+q}\binom{k}{q}\binom{\ell}{k} = \left\{\begin{array}{ccc} 1 \mbox{ if } \ell =q\\ 0 \mbox{ if }\ell >q\end{array}\right.
I don't see what to do here. Can we interpret it combinatorially somehow?
Answer
HINT: Assume that \ell>q, since the other case is trivial. Suppose that we have white slips of paper numbered 1 through \ell. If k\le q\le\ell, \binom{\ell}k\binom{k}q is the number of ways to choose k of those slips and color them blue, then choose q of the k blue slips and circle the number. This is clearly the same as choosing q of the original \ell slips, coloring them blue, and circling the number, and then choosing k-q of the remaining \ell-q white slips and coloring them blue, so \binom{\ell}k\binom{k}q=\binom{\ell}q\binom{\ell-q}{k-q}. And k+q\equiv k-q\pmod2, so, setting j=k-q,
\sum_{k=q}^\ell(-1)^{k+q}\binom{\ell}k\binom{k}q=\binom{\ell}q\sum_{j=0}^{\ell-q}(-1)^j\binom{\ell-q}j\;.
Can you finish it from there?
No comments:
Post a Comment