Question

Evaluate 9p3 and 10c7

Answer 1

Permutations of n things taken r at a time is computed as follows:

$\text{nPr}=\frac{n!}{(n-r)!}$

Replacing with n = 9 and r = 3,

$9P3=\frac{9!}{(9-3)!}=\frac{362880}{720}=504$

Combinations of n things taken r at a time is computed as follows:

$nCr=\frac{n!_{}}{(n-r)!\cdot r!}$

Replacing with n = 10 and r = 7,

$10C7=\frac{10!}{(10-7)!\cdot7!}=\frac{3628800}{6\cdot5040}=120$