AP EAMCET · Maths · Permutation Combination
The number of ways of dividing 15 persons into 3 groups containing 3,5 and 7 persons so that two particular persons are not included into the 5 persons group is
- A \(\frac{117(11!)}{3!(7!)}\)
- B \({ }^{15} \mathrm{C}_5{ }^{10} \mathrm{C}_3\)
- C \(90 \times \frac{13!}{7!}\)
- D \({ }^{15} \mathrm{C}_5{ }^8 \mathrm{C}_3\)
Answer & Solution
Correct Answer
(A) \(\frac{117(11!)}{3!(7!)}\)
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