Permutations And Combinations

Question
CBSEENMA11014112

In how many ways can 6-girls and 4 boys be seated in a row so that no two boys are together?

Solution

No two boys are together.
Step I:  Arrange 6 girls in a row by leaving one seat between every two girls.
            Number of permutations of arranging girls = 6! = 720            ...(i)
      circled times  G  circled times  G  circled times  G  circled times  G  circled times  G  circled times  G     circled times
       1         2        3          4      5        6             7     
Step II:  There are 7 places for the boys to sit in order that no two of them are together.
                     Number of boys = 4
              rightwards double arrow              n = 7,  r = 4
∴             Number of permutations of arranging boys = straight P presuperscript 7 subscript 4 space equals space fraction numerator 7 factorial over denominator 3 factorial end fraction space equals space 7 cross times 6 cross times 5 cross times 4 space equals space 840 ...(ii)
                Hence, from (i) and (ii),
                 The number of ways in which they may be seated so that no two boys are together
                                                       = 720 x 840 = 604800

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