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AP EAMCET · Maths · Permutation Combination

If \({ }^{27} \mathrm{P}_{\mathrm{r}+7}=7722{ }^{25} \mathrm{P}_{(\mathrm{r}+4)}\), then \(\mathrm{r}=\)

  1. A 9
  2. B 12
  3. C 11
  4. D 10
Verified Solution

Answer & Solution

Correct Answer

(D) 10

Step-by-step Solution

Detailed explanation

\(\frac{27!}{(27 - (r+7))!} = 7722 \frac{25!}{(25 - (r+4))!}\) \(\frac{27!}{(20 - r)!} = 7722 \frac{25!}{(21 - r)!}\) \(27 \times 26 = \frac{7722}{21 - r}\) \(702 = \frac{7722}{21 - r}\) \(21 - r = \frac{7722}{702}\) \(21 - r = 11\) \(r = 21 - 11\) \(r = 10\)