5 girls and 10 boys sit at random in a row having 15 chairs numbered as 1 to 15, then find the probability that end seats are occupied by the girls and between any two girls an odd number of boys sit
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Sol. 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 ⇒ 2k 1 + 1 + 2k 2 + 1 + 2k 3 + 1 + 2k 4 + 1 = 10
⇒ k 1 + k 2 + k 3 + k 4 = 3 where k i ≥ 0
number of solution are 3+ 4–1 C 4–1 = 6 C 3 ⇒ n (E) = 6 C 3 × 10! × 5! ⇒ Now P = 
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