# Friday Morning Math Problem

## A Binomial Problem

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

Find r.

Challenge problem:

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

Find r.

Solution
First, if we solve it naively, we get

r = 2r + 1

-r = 1

r = -1

We use the fact that ${\displaystyle {\binom {n}{m}}={\binom {n}{n-m}}}$

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 < s <= n such that ${\displaystyle {\binom {n}{r}}={\binom {n}{s}}}$

We can prove that ${\displaystyle {\binom {n}{r}}={\binom {n}{s}}}$ only holds if r = s or r = n - s

This is something we may implicitly assume, due to our own understanding of the nature of the binomial numbers and Pascal's Triangle. But to be rigorous we have to cover all our bases.

Challenge problem:

Solving naively we get

r = 3r + 4

-4r = 4

r = -1

To transform, use the same identity as above

3r + 4 = r

3r + 4 = 120 - r

4r = 120 - 4

r = (120 - 4)/4 = 29