Question

Evaluate 9p3 and 10c7

Answer 1

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

`\text{nPr}=(n!)/((n-r)!)`

Replacing with n = 9 and r = 3,

`9P3=(9!)/((9-3)!)=(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=(10!)/((10-7)!\cdot7!)=(3628800)/(6\cdot5040)=120`