If `m` things are distributed among `a` men and `b` women, show that the chance that the number of things received by man is `1/2(((b+a)^m-(b-a)^m)/((b+a)^m))dot`

Explanation

Suppose than p and q, respectively, denote the probabilities that a thing goes to a man and a women.
`p=(a)/(a+b)and q=(b)/(a+b)`
Now, the probabilities of 0,1,2,3,… thing going to men are the first, second third terms etc. in the following binomial expansion. `(q+p)^(m)=q^(m)+""^(m)C_(1)a^(m-2)p^(2)+...+p^(m)" "(1)`
But men aer to receive and odd number of things.
Hence the required probability is the sum of even terms in the equation (1).
To obtain the sum of evetn terms, we write the expansion
`(q-p)^(m)=q^(m)-""^(m)C_(1)a^(m-1)p+""^(m)C_(2)a^(m-2)p^(2)-...+(-1)^(m)p^(m)" "(2)`
Subtracting (2) from (1), we get
`(q+p)^(m)-(q-p)^(m)=` Sum of even terms in (1)
Hence, the required probability
`=1/2[(q+p)^(m)-(q-p)^(m)]`
`=1/2[1-((b-a)/(b+a))^(m)]`
`=((b+a)^(m)-(b-a)^(m))/(2(b+a)^(m))`