AP EAMCET · Maths · Permutation Combination
If \({ }^{27} \mathrm{P}_{\mathrm{r}+7}=7722{ }^{25} \mathrm{P}_{(\mathrm{r}+4)}\), then \(\mathrm{r}=\)
- A 9
- B 12
- C 11
- D 10
Answer & Solution
Correct Answer
(D) 10
Step-by-step Solution
Detailed explanation
\(\frac{27!}{(27 - (r+7))!} = 7722 \frac{25!}{(25 - (r+4))!}\) \(\frac{27!}{(20 - r)!} = 7722 \frac{25!}{(21 - r)!}\) \(27 \times 26 = \frac{7722}{21 - r}\) \(702 = \frac{7722}{21 - r}\) \(21 - r = \frac{7722}{702}\) \(21 - r = 11\) \(r = 21 - 11\) \(r = 10\)
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