Q48 / 160 · TS EAMCET 2008 · Mathematics

If ${ }^n P_r=30240$ and ${ }^n C_r=252$, then the ordered pair $(n, r)$ is equal to

  • A
    $(12,6)$
  • Correct answer
    B
    $(10,5)$
  • C
    $(9,4)$
  • D
    $(16,7)$

Solution

Given that, ${ }^n P_r=30240$ and ${ }^n C_r=252$ $$ \begin{aligned} & \Rightarrow \frac{n !}{(n-r) !}=30240 \text { and } \frac{n !}{(n-r) ! r !}=252 \\ & \Rightarrow \quad r !=\frac{30240}{252}=120 \\ & \Rightarrow \quad r=5 \\ & \therefore \quad \frac{n !}{(n-5) !}=30240 \\ & \end{aligned} $$ $$ \begin{aligned} & \Rightarrow \quad n(n-1)(n-2)(n-3)(n-4)=30240 \\ & \Rightarrow \quad n(n-1)(n-2)(n-3)(n-4) \\ & \quad=10(10-1)(10-2)(10-3)(10-4) \\ & \Rightarrow \quad n=10 \end{aligned} $$ Hence, required ordered pair is $(10,5)$.

Hence, the correct answer is option: 'B'.

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