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Question

5 girls and 10 boys sit at random in a row having 15 chairs numbered as 1 to 15. The probability that end seats are occupied by the girls and between any two girls odd number of boys sit, is

A
20×10!×5!15!
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B
20×10!15!
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C
20×5!15!
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D
None of these
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Solution

The correct option is A 20×10!×5!15!
There are four gaps in between the girls where the boys can sit.
Let the number of boys in these gaps be 2a+1,2b+1,2c+1,2d+1=0.
Then 2a+1+2b+1+2c+1+2d+1=10a+b+c+d=3
The number of solutions of above equation = coefficients of x3 in (1x)4=6C3=20
Thus, boys and girls can sit in 20×10!×5! ways
Total ways =15!
Hence, the required probability =20×10!×5!15!

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