A composition of the number $n$ with $k$ summands is the representation $ n = a_1 + \cdots + a_k$ with integers $a_i \geq 1, 1 \leq i \leq k$. The order of the summands is important.
Prove: There are as many compositions of n with k even summands as compositions of n-k with k odd summands exist.
Hi!
I don't have any idea how to prove that as I don't know how to handle this "odd" and "even" stuff. Bijection? Induction?
Could you please help me to understand this?
Thanks in advance!