# Partitions proof

• May 14th 2012, 04:14 PM
Partitions proof
Let the number of partitions of n into parts each less than or equal to m be denoted P(n,m). Prove
P(n,m) = P(n, m-1) + P(n-m,m)
• May 14th 2012, 04:40 PM
Plato
Re: Partitions proof
Quote:

Let the number of partitions of n into parts each less than or equal to m be denoted P(n,m). Prove
P(n,m) = P(n, m-1) + P(n-m,m)

I have written on this question. So I tell you whoever wrote your question has left out some cases.
What happens if $m=1~?$ What happens if $m=n~?$

Here is the MathCad way of doing it.
[ATTACH=CONFIG]23869[/ATTACH]
• May 14th 2012, 07:46 PM
Re: Partitions proof
what is mathcad? >.> ... i'm not really following that equation :(
I'm a little confused because there is a formula in the book about partitioning n into exactly r parts where order counts = C(n-1,r-1)
but then there's another one that says "the number of ways of writing n as the sum of r or fewer integers where the order of the summands matters is C(r+n-1,n)"
but i didn't think they applied because order shouldn't count, should it?
I was also trying to play with the concept that the number of partitions of n into r or less parts = the number of partitions of n into parts that are r or less, but have gotten nowhere so far :\

Quote:

Originally Posted by Plato

if m = 1 there should only be one partition, n itself, right?
and if m = n, then all the partitions = 1
• May 17th 2012, 09:10 PM