AP EAMCET · Maths · Permutation Combination
5 boys and 6 girls are arranged in all possible ways. Let X denote the number of linear arrangements in which no two boys sit together and \(Y\) denote the number of linear arrangements in which no two girls sit together. If \(Z\) denote the number of ways of arranging all of them around a circular table such that no two boys sit together, then \(X: Y: Z=\)
- A \(1: 1: 21\)
- B \(21: 1: 1\)
- C \(7: 5: 5\)
- D \(4: 3: 3\)
Answer & Solution
Correct Answer
(B) \(21: 1: 1\)
Step-by-step Solution
Detailed explanation
When no two boys sit together, \(* \mathrm{G} * \mathrm{G} * \mathrm{G} * \mathrm{G} * \mathrm{G} * \mathrm{G} *\) \(\therefore X=\) Number of ways of arrangement in which no two boys sit together \(={ }^7 C_5 \times 5!\times 6!\) When no two girls sit together…
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