Question

How many combinations are possible without repeats for C(6,2)

Answer 1

The combinations can be calculated using the following formula:

`C(n,r)=(n!)/(r!(n-r)!)`

In our case n=6 and r=2, then:

`C(6,2)=(6!)/(2!(4)!)=(2*3*4*5*6)/(2*2*3*4)=(5*6)/(2)=15`

Hence the answer is 15