Question

If 8Cr-7C3=7C2 find r

Answer 1

given, $^8C_r-^7C_3=^7C_2$

we know, $^nC_r=\frac{n!}{r!(n-r)!}$
use this basic formula to get r,

$^8C_r=\frac{8!}{r!(8-r)!}\\\\\\^7C_3=\frac{7!}{3!.4!}\\\\\\^7C_2=\frac{7!}{2!.5!}$

$\frac{8!}{r!(8-r)!}-\frac{7!}{3!.4!}=\frac{7!}{2!.5!}$

$\frac{8!}{r!(8-r)!}=\frac{7!}{3!.4!}+\frac{7!}{2!.5!}$

$\frac{8!}{r!(8-r)!}=7!\left(\frac{1}{3!.4!}+\frac{1}{2!.5!}\right)$

$\frac{8!}{r!(8-r)!}=7!\left(\frac{5}{3.2!5!}+\frac{1}{2!.5!}\right)$

$\frac{8!}{r!(8-r)!}=7!\left(\frac{8}{3.2!.5!}\right)$

$\frac{8!}{r!(8-r)!}=\frac{8.7!}{3!.5!}$

$\frac{8!}{r!.(8-r)!}=\frac{8!}{3!.(8-3)!}$

compare LHS to RHS
so, r = 3