Question

Prove the relation rC(n r)=nC(n-1 r-1)

Answer 1

C(n,r)=C(n−1,r−1)+C(n−1,r)

Step-by-step explanation:

Correct Question is

C(n,r)=C(n−1,r−1)+C(n−1,r)

replacing r with a

ⁿCₐ = ⁿ⁻¹Cₐ₋₁ + ⁿ⁻¹Cₐ

RHS =  ⁿ⁻¹Cₐ₋₁ + ⁿ⁻¹Cₐ

= (n-1)!/(a - 1)!(n-1 -a + 1)!   + (n-1)!/a!(n-1 - a)!

= (n-1)!/(a - 1)!(n -a)!   + (n-1)!/a!(n-1 - a)!

= (n-1)!/(a - 1)!(n -a)(n -a-1)!  + (n-1)!/a(a - 1)!(n-a- 1)!

= ((n - 1)!/(a - 1)!(n-a- 1)!) ( 1/(n -a)  +  1/a)

= ((n - 1)!/(a - 1)!(n-a- 1)!) (( a + n - a)/(a)(n-a))

= ((n - 1)!/(a - 1)!(n-a- 1)!) (n/(a)(n-a))

= (n(n - 1)! / a(a-1)! (n-a)(n-a- 1)!

= n!/a!(n-a)!

= ⁿCₐ

= LHS

QED

proved

Hence ⁿCₐ = ⁿ⁻¹Cₐ₋₁ + ⁿ⁻¹Cₐ

C(n,r)=C(n−1,r−1)+C(n−1,r)

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