`|[n, n+1, n+2] , [P(n,n) , P(n+1, n+1), P(n+2, n+2)] , [ C(n,n), C(n+1, n+1), C(n+2, n+2)]|=`

If n C_(1)+2. n C_(2)+..n. n C_(n) = 2 n^(2) , then n =

Given that C(n, r) : C(n,r + 1) = 1 : 2 and C(n,r + 1) : C(n,r + 2) = 2 : 3 . What is P(n, r) : C(n, r) equal to ?

Let f (n) = |{:(n,,n+1,,n+2),(.^(n)P_(n),,.^(n+1)P_(n+1),,.^(n+2)P_(n+2)),(.^(n)C_(n),,.^(n+1)C_(n+1),,.^(n+2)C_(n+2)):}| where the sysmbols have their usual neanings .then f(n) is divisible by

Let f (n) = |{:(n,,n+1,,n+1),(.^(n)P_(n),,.^(n+1)P_(n+1),,.^(n+2)P_(n+2)),(.^(n)C_(n),,.^(n+1)C_(n+1),,.^(n+2)C_(n+2)):}| where the sysmbols have their usual neanings .then f(n) is divisible by

Let f (n) = |{:(n,,n+1,,n+1),(.^(n)P_(n),,.^(n+1)P_(n+1),,.^(n+2)P_(n+2)),(.^(n)C_(n),,.^(n+1)C_(n+1),,.^(n+2)C_(n+2)):}| where the sysmbols have their usual neanings .then f(n) is divisible by

Find n if : ""^(2n+1) P_(n-1) : ""^(2n-1) P_n = 3: 5

If .^(2n+1)P_(n-1): .^(2n-1)P_(n)=3:5 , find n.

Let f(n)=|n n+1 n+2 ^n P_n^(n+1)P_(n+1)^n P_n^n C_n^(n+1)C_(n+1)^(n+2)C_(n+2)| , where the symbols have their usual meanings. Then f(n) is divisible by a. n^2+n+1 b. (n+1)! c. n ! d. none of these

""^(2n)C_(n+1)+2. ""^(2n)C_(n) + ""^(2n) C_(n-1) =