JEE Mains · Maths · STD 11 - 6. permutation and combination
The number of ways of forming a queue of \(4\) boys and \(3\) girls such that all the girls are not together, is:
- A \(5040\)
- B \(3050\)
- C \(3410\)
- D \(4320\)
Answer & Solution
Correct Answer
(D) \(4320\)
Step-by-step Solution
Detailed explanation
Total number of boys \(= 4\) Total number of girls \(= 3\) Total number of children \(= 7\) Total number of ways to arrange \(7\) children in a queue \(= 7! = 5040\) To find the number of ways where all \(3\) girls are together, consider the \(3\) girls as a single unit. Total…
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