`sum_(p=1)^(n) sum_(m=p)^(n) (("n"),("m"))(("m"),("p")) ` is equal to

Options

A. (a)`3^(n)`
B. (b)`2^(n) `
C. (c)`3^(2) + 2^(n)`
D. (d)`3^(n) - 2^(n)`

Answer

d

Explanation

`therefore ((n),(n))((m),(p)) = (n!)/(n!(n-m)!) xx(m!)/(p!(m-p)!)`
` =(n!)/((n-m)!p!(m-p)!) = ((n),(p)) ((n-p),(m-p))`
`therefore sum_(p=1)^(n) sum_(m=p)^(n) ((n),(m))((m),(p)) = sum_(p=1)^(n) sum_(m=p)^(n) ((n),(p))((n-p),(m-p)) `
` = sum_(p=1)^(n) ((n),(p))sum_(m=p)^(n)((m-p),(m-p))`
` sum_(p-1)^(n) ((n),(p)) sum_(t=0)^(n)((n-p),(t)) "[where ", t= m-p]`
` = 2^(n) sum_(p=1)^(n) ((n),(p)) 2^(n-p)`
` = 2^(n) sum_(p=1)^(n) ((n),(p))(1)/( 2^(p)) = 2^(n) [(1+ (1)/(2))^(n)-1] = 3^(n) - 2^(n)`