04 Sep '17 18:202 edits

why does

∑[g=h, ∞] ( C(h – 1, m – 1) / C(g, m) ) = m/(m – 1)

where

C(n, k) = n! / ( k! (n – k)! ) and is a binomial coefficient.

g, h, m ∈ ℕ

g≥h≥1, h≥m≥1

?

I checked this with a computer program and it appears to be always correct but cannot see algebraically why it should be.

Using C(n, k) = n! / ( k! (n – k)! )

that can be reexpressed as

∑[g=h, ∞] ( (h – 1)! /( (m – 1)! (h – m)! ) / ( g! / ( m! (g – m)! ) )

= ∑[g=h, ∞] m! (g – m)! (h – 1)! / ( g! (m – 1)! (h – m)! )

but

m! / (m – 1)! = m

thus

= ∑[g=h, ∞] m (g – m)! (h – 1)! / ( g! (h – m)! )

= m * ∑[g=h, ∞] (g – m)! (h – 1)! / ( g! (h – m)! )

but then why does

∑[g=h, ∞] (g – m)! (h – 1)! / ( g! (h – m)! ) = 1/(m – 1)

where

g, h, m ∈ ℕ

g≥h≥1, h≥m≥1

?

∑[g=h, ∞] ( C(h – 1, m – 1) / C(g, m) ) = m/(m – 1)

where

C(n, k) = n! / ( k! (n – k)! ) and is a binomial coefficient.

g, h, m ∈ ℕ

g≥h≥1, h≥m≥1

?

I checked this with a computer program and it appears to be always correct but cannot see algebraically why it should be.

Using C(n, k) = n! / ( k! (n – k)! )

that can be reexpressed as

∑[g=h, ∞] ( (h – 1)! /( (m – 1)! (h – m)! ) / ( g! / ( m! (g – m)! ) )

= ∑[g=h, ∞] m! (g – m)! (h – 1)! / ( g! (m – 1)! (h – m)! )

but

m! / (m – 1)! = m

thus

= ∑[g=h, ∞] m (g – m)! (h – 1)! / ( g! (h – m)! )

= m * ∑[g=h, ∞] (g – m)! (h – 1)! / ( g! (h – m)! )

but then why does

∑[g=h, ∞] (g – m)! (h – 1)! / ( g! (h – m)! ) = 1/(m – 1)

where

g, h, m ∈ ℕ

g≥h≥1, h≥m≥1

?