Year 12s: what will be the first way you research your future options? >>

Binomial relation

Can someone please explain to me how

\displaystyle k\binom{p}{k} = p\binom{p - 1}{k - 1}

Where p is prime and k is a positive integer less than p.

?

EDIT: Nevermind... I kept using the formula for permutations and not combinations :facepalm:

(edited 10 years ago)

Reply 1

mathmari

k\binom{p}{k}=k\frac{p!}{k!(p-k)!}=\frac{p!}{(k-1)!(p-k)!}=\frac{p(p-1)!}{(k-1)!(p-k-1+1)!}=p\frac{(p-1)!}{(k-1)!((p-1)-(k-1))!}=p\binom{p-1}{k-1}

Reply 2

0x2a

Original post by mathmari

k\binom{p}{k}=k\frac{p!}{k!(p-k)!}=\frac{p!}{(k-1)!(p-k)!}=\frac{p(p-1)!}{(k-1)!(p-k-1+1)!}=p\frac{(p-1)!}{(k-1)!((p-1)-(k-1))!}=p\binom{p-1}{k-1}

Yep, I kept using the formula for permutations, silly me :colondollar:

Thanks anyways :biggrin:

Quick Reply

Latest

Articles for you

Why you need to know about contextual offers before you apply to university

How to revise for GCSE Sociology exams: AQA explains what to do

How to revise for GCSE Design Technology exams: AQA explains what to do

Terms and conditions

Binomial relation

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you