Maths · Simple applications
7 boys and 8 girls have to sit in a row on 15 chairs numbered from 1 to 15 then?
7 boys and 8 girls have to sit in a row on 15 chairs numbered from 1 to 15 then?
- A. Number of ways boys and girls sit alternately is \( 8 ! 7 ! \)
- B. Number of ways boys and girls sit alternately is \( 2(8 ! 7 !) \)
- C. The number of ways in which first and fifteenth chair are occupied by boys and between any two boys an even number of girls sit is \( ^{9} C_{4} 8 ! 7 ! \)
- D. The number of ways in which first and last seat are occupied by boys and between any two boys an even number of girls sit is \( \left(2^{9} C_{4} 8 ! 7 !\right) \)
Step-by-step solution
For the condition, the number of boys is 7, so there are 6 gaps between them. Let the number of girls in each gap be 2y_i, then sum y_i = 4. Number of nonnegative integer solutions is C(4+6-1,6-1)=C(9,5)=126. Thus total ways = 126 * 7! * 8! = 9C4 * 8! * 7!.
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