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Home  >>  CBSE XI  >>  Math  >>  Permutations and Combinations

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

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

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1 Answer

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.
answered May 16, 2014 by sreemathi.v
 

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