TS EAMCET · Maths · Permutation Combination
10 men and 6 women are to be seated in a row so that no two women sit together. The number of ways they can be seated, is
- A \(11 ! 10 !\)
- B \(\frac{11 !}{6 ! 5 !}\)
- C \(\frac{10 ! 9 !}{5 !}\)
- D \(\frac{11 ! 10 !}{5 !}\)
Answer & Solution
Correct Answer
(D) \(\frac{11 ! 10 !}{5 !}\)
Step-by-step Solution
Detailed explanation
W W W W W W W W W First, we arrange 10 men in a row at alternate position. So, number of ways formula \(=10\) ! Now, 6 women can arrange in 11 positions So, number of ways for women \(={ }^{11} P_6\) Required number of…
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