Question

3. C(8, r) = 28; r =​

Answer 1

Given:

$C(8,r)=28$

To find:

The value of r.

Solution:

Combination formula:

$C(n,r)=\dfrac{n!}{r!(n-r)!}$

We have,

$C(8,r)=28$

Using the combination formula, we get

$\dfrac{8!}{r!(8-r)!}=28$

$\dfrac{8!}{28}=r!(8-r)!$

$\dfrac{8\times 7\times 6!}{28}=r!(8-r)!$

$2\times 6!=r!(8-r)!$

It can be written as:

$2!6!=r!(8-r)!$                $[\because 2!=2\times 1=2]$

Case 1:

$2!=r!$

$r=2$

And,

$6!=(8-r)!$

$6=8-r$

$r=8-6$

$r=2$

Case 2:

$6!=r!$

$r=6$

And,

$2!=(8-r)!$

$2=8-r$

$r=8-2$

$r=6$

Therefore, the value of r is either 2 or 6.