How Do You Simplify Logarithmic Expressions in Combinatorics?

[Jenkz]

Homework Statement

Show that ln[ (N + M - 1)! /M! (N-1)! ] is equal to N ln((N+M) / N) + M ln((N+M) /M).

Homework Equations

Using stirling's formula ln N! ~ N lnN - N

The Attempt at a Solution

ln[ (N + M - 1)! /M! (N-1)! ] (a)
= (N+M -1) ln(N+M -1)- (N+M -1) - M lnM + M - (N-1) ln(N-1) + (N-1) (b)
=(N+M) ln(N+M) - M lnM - N lnN (c)
= N ln( (N+M)/N) + M ln ( (N+M) / M) (d)I understand how to get from (a) to (b), and (c) to (d). But I don't understand what happens from (b) to (c). What has happened to the -1 values?

Thanks.

 

[mighty2000]

In step (b), the -1 values have been factored out. This is because of the properties of logarithms, specifically the property that ln(a/b) = ln(a) - ln(b). So in step (b), the -1 values have been factored out and combined with the -1 in the numerator, resulting in (N+M-1). Similarly, the (N-1) term in the denominator has been factored out and combined with the (N-1) in the numerator, resulting in just N. This is why in step (c), the -1 values are no longer present.