Question

Ten individuals are candidates for a committee. Two will be selected. How many different pairs of individuals can be selected?

Answer 1

Answer:

45

Step-by-step explanation:

The different pairs of individual can be computed using rule of combination.

The combination is denotes as nCr.

nCr=n!/(r!(n-r)!)

In the given problem n=total candidates=10 and r= Selected candidates=2.

10C2=10!/(2!(10-2)!)

10C2=10*9*8!/(2!8!)

10C2=90/2

10C2=45

So, 45 different pairs of individuals can be selected.

Answer 2

Answer:

45

Step-by-step explanation:

Pairs that can be selected = 10C2

$= \frac{10!}{2!(10-2)!}= \frac{10!}{2!*8!}\\ \\= \frac{10*9*8!}{2*1*8!}= \frac{10*9}{2*1}\\ \\= 5 * 9 = 45$