Question

8Cr-7C3=7C2 find the value of r

Answer 1

Check this attachment :-)

Answer 2

Answer:

The value of r=3.          

Step-by-step explanation:

Given : $^8C_r-^7C_3=^7C_2$

To find : The value of r ?

Solution :

We know that,

$^nC_r=\frac{n!}{r!(n-r)!}$

$^8C_r-^7C_3^7C_2$

Applying formula,

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

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

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

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

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

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

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

$\frac{1}{r!(8-r)!}=\frac{1}{5!3!}$

When we compare and solve, r=3