Question: 2.6 Problems 37 where m n N a + m. Hence, show that a N a 1 m 1
2.6 Problems 37 where m ≤ n ≤ N − a + m. Hence, show that a
N
a − 1 m − 1
(N − a)!
(N − 1)!
N−Xa+m n=m
(n − 1)! (N − n)!
(n − m)! (N − a + m − n)! = 1, and that the expectation of n is N + 1 a + 1 m. (Oxford 1972M)
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