Question

Prove that
n-1c3+n-1c4>nc3

Answer 1

according to the formula of combination ,
$ncr + nc(r - 1) =( n + 1)cr$

Answer 2

This can be solved using the formulas of PNC

ⁿ⁻¹C₃  can be written as :

(n-1)!/3! (n-1-3 )! = (n-1)!/3! * (n-4) !

= (n-1)(n-2)(n-3)(n-4)!/6*( n-4)!

= (n-1)(n-2)(n-3)/6

ⁿ⁻¹C₄ can be written as :

(n-1)!/4! (n-1-4 )! = (n-1)!/4! * (n-5) !

= (n-1)(n-2)(n-3)(n-4)(n-5)!/24*( n-5)!

= (n-1)(n-2)(n-3)(n-4) /24

Adding these two:

=(n-1)(n-2)(n-3)/6 +  (n-1)(n-2)(n-3)(n-4) /24

=  (n-1)(n-2)(n-3)/6 (  1+ (n-4)/4)

= n(n-1)(n-2)(n-3)/24

= n(n-1)(n-2)/6 *( (n-3)/4)

From RHS, ⁿC₃ can be written as:

= n!/3!*(n-3)!

= n(n-1)(n-2)(n-3)!/3!*(n-3)!

= n(n-1)(n-2)/6

Hence, RHS would always be less than the LHS of the equation