ProbabilityHard
Question
5 girls and 10 boys sit at random in a row having 15 chairs numbered as 1 to 15, then the probability that end seats are occupied by the girls and between any two girls an odd number of boys sit is:
Options
A.
B.
C.
D.
Solution
n(S) = ways of sitting of 10 boys and 5 girls = 15!
Let end seats are occupied by the girls & between first and second girl x boys are seated similarly between second and third y boys ............. so on then
x + y + z + w = 10
where x, y, z, w are (2k + 1) type
2k1 + 1 + 2k2 + 1 + 2k3 + 1 + 2k4 + 1 = 10
⇒ k1 + k2 + k3 + k4 = 3 where ki ≥ 0
number of solution are 3 + 4 - 1C4 - 1 = 6C3
n (E) = 6C3 ≠ 10! × 5!
Let end seats are occupied by the girls & between first and second girl x boys are seated similarly between second and third y boys ............. so on then
x + y + z + w = 10
where x, y, z, w are (2k + 1) type
2k1 + 1 + 2k2 + 1 + 2k3 + 1 + 2k4 + 1 = 10
⇒ k1 + k2 + k3 + k4 = 3 where ki ≥ 0
number of solution are 3 + 4 - 1C4 - 1 = 6C3
n (E) = 6C3 ≠ 10! × 5!
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