Given:
[tex]C(8,r)=28[/tex]
To find:
The value of r.
Solution:
Combination formula:
[tex]C(n,r)=\dfrac{n!}{r!(n-r)!}[/tex]
We have,
[tex]C(8,r)=28[/tex]
Using the combination formula, we get
[tex]\dfrac{8!}{r!(8-r)!}=28[/tex]
[tex]\dfrac{8!}{28}=r!(8-r)![/tex]
[tex]\dfrac{8\times 7\times 6!}{28}=r!(8-r)![/tex]
[tex]2\times 6!=r!(8-r)![/tex]
It can be written as:
[tex]2!6!=r!(8-r)![/tex] [tex][\because 2!=2\times 1=2][/tex]
Case 1:
[tex]2!=r![/tex]
[tex]r=2[/tex]
And,
[tex]6!=(8-r)![/tex]
[tex]6=8-r[/tex]
[tex]r=8-6[/tex]
[tex]r=2[/tex]
Case 2:
[tex]6!=r![/tex]
[tex]r=6[/tex]
And,
[tex]2!=(8-r)![/tex]
[tex]2=8-r[/tex]
[tex]r=8-2[/tex]
[tex]r=6[/tex]
Therefore, the value of r is either 2 or 6.
