FMM17

Friday Morning Math Problem

A Binomial Problem

$ \binom{16}{r} = \binom{16}{2r+1} $

Find r.

Challenge problem:

$ \binom{120}{r} = \binom{120}{3r+4} $

Find r.

Solution

First, if we solve it naively, we get r = 2r + 1 -r = 1 r = -1 We use the fact that $ \binom{n}{m} = \binom{n}{n-m} $ 2r + 1 =16 - (2r+1) r = 16 - (2r+1) r = 16 - 2r - 1 3r = 15 r = 5 What more do we need to do? We can use this identity to transform r = 2r + 1, but there may also exist other integers 0 <= r < l <= n such that $ \binom{n}{r} = \binom{n}{l} $ We can prove that $ \binom{n}{r} = \binom{n}{l} $ only holds if r = l or r = n - l This is something we may implicitly assume, due to our own understanding of the nature of the binomial numbers and Pascal's Triangle. We just have to cover all our bases. Challenge problem: r = (120 - 4)/(1 + 3) = 29