Question

evaluate 11p4 and 9c5

Answer 1

$\bf _{n}P_{r}=\cfrac{n!}{(n-r)!}\qquad \qquad _{11}P_4=\cfrac{11!}{(11-4)!} \\\\\\ _nC_r=\cfrac{n!}{r!(n-r)!}\qquad \qquad _9C_5=\cfrac{9!}{5!(9-5)!}$


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