Question

Compute the permutations and combinations. 4 C 2

Answer 1

$4C2= \frac{4!}{2!(4-2)!}= \frac{4*3*2*1}{(2*1)(2!)}= \frac{4*3*2*1}{(2*1)(2*1)}=6$

I hope that helps!

Answer 2

Answer:

Combination:

$C(4,2)=4C_2 =6$

Permutation:

$P(4,2)=4P_2=12$

Step-by-step explanation:

A permutation of a set of elements is an arrangement of said elements taking into account the order. A combination of a set of elements is a selection of those elements regardless of order.

The number of permutations "k" of "n" elements  is calculated with the following formula:

$P(n,k)=nP_k =\frac{n!}{(n-k)!}$

The number of combinations "k" of "n" elements is calculated with the following formula:

$C(n,k)=nC_k=\frac{n!}{k!(n-k)!}$

Therefore:

$C(4,2)=4C_2=\frac{4!}{2!(4-2)!} =\frac{24}{2(2)}=\frac{24}{4} =6$

and

$P(4,2)=4P_2=\frac{4!}{(4-2)!} =\frac{24}{2} =12$