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# If $nP_r=840,nC_r=35$ then $r$=_______

$\begin{array}{1 1}(A)\;6\\(B)\;4\\(C)\;8\\(D)\;9\end{array}$

Toolbox:
• $nP_r=\large\frac{n!}{(n-r)!}$
• $nC_r=\large\frac{n!}{r!(n-r)!}$
Given : $nP_r=840$
$\large\frac{n!}{(n-r)!}=$$840-----(1) nC_r=35 \large\frac{n!}{r!(n-r)!}$$=35$
Equating equation (1) and (2) we get
$nC_r=\large\frac{1}{r!}$$nP_r 35=\large\frac{1}{r!}$$840$
$r!=\large\frac{840}{35}$
$r!=24$
$r=4$
Hence (B) is the correct answer.