Prove that the number of ways to select `n` objects from `3n` objects of whilch `n` are identical and the rest are different is `2^(2n-1)+((2n) !)/(2(n !)^2)`

If n is a natural number then show that 1.3.5.7…..(2n-1) = ((2n)!)/(2^n n!)

If N denotes the number of ways of selecting r objects of out of n distinct objects (rgeqn) with unlimited repetition but with each object included at least once in selection, then N is equal is a. .^(r-1)C_(r-n) b. .^(r-1)C_n c. .^(r-1)C_(n-1) d. none of these

If n objects are arrange in a row,then the number of ways of selecting three of these objects so that no two of them are next to each other is a.^(n-2)C_(3) b.^(n-3)C_(2) c.^(n-3)C_(3) d.none of these

Prove that n! (n+2) = n! +(n+1)! .

Prove that 2n +1 lt 2^(n) for all natural numbers n ge3

Prove that (2^(n)+2^(n-1))/(2^(n+1)-2^(n))=(3)/(2)

The total number of ways of selecting two numbers from the set {1,2,3,4,........3n} so that their sum is divisible by 3 is equal to a.(2n^(2)-n)/(2) b.(3n^(2)-n)/(2) c.2n^(2)-n d.3n^(2)-n