value of`1*^(30)C_(0)*^(20)C_(0)+2.^(30)C_(1)*^(20)C_(1)+3*^(30)C_(2)*^(20)C_(2)+....+21.^(20)C_(20)*^(30)C_(20)` `is*^(50)C_(20)+20.^(49)C_(q) `(where q is even number) then q is equal to

"^(30)C_(0)*^(20)C_(10)+^(31)C_(1)*^(19)C_(10)+^(32)C_(2)*18C_(10)+....^(40)C_(10)*^(10)C_(10) is equal to

The value of .^(20)C_(0)+.^(20)C_(1)+.^(20)C_(2)+.^(20)C_(3)+.^(20)C_(4)+.^(20)C_(12)+.^(20)C_(13)+.^(20)C_(14)+.^(20)C_(15) is

Evaluate: .^(20)C_(5)+^(20)C_(4)+^(21)C_(4)+^(22)C_(4)

The sum of the series .^(20)C_(0)-.^(20)C_(1)+ .^(20)C_(2)-.^(20)C_(3)+...-.+ .^(20)C_(10) is -

Find the value of .^(20)C_(0) xx .^(13)C_(10) - .^(20)C_(1) xx .^(12)C_(9) + .^(20)C_(2) xx .^(11)C_(8) - "……" + .^(20)C_(10) .

If a= .^(20)C_(0) + .^(20)C_(3) + .^(20)C_(6) + .^(20)C_(9) + "…..", b = .^(20)C_(1) + .^(20)C_(4) + .^(20)C_(7) + "……"' and c = .^(20)C_(2) + .^(20)C_(5) + .^(20)C_(8) + "…..", then Value of a^(3) + b^(3) + c^(3) - 3abc is

The value of r for which .^(20)C_(r ), .^(20)C_(r - 1) .^(20)C_(1) + .^(20)C_(2) + …… + .^(20)C_(0) .^(20)C_(r ) is maximum, is

The value of r for which .^(20)C_(r ), .^(20)C_(r - 1) .^(20)C_(1) + .^(20)C_(2) + …… + .^(20)C_(0) .^(20)C_(r ) is maximum, is

If .^(20)C_(1)+(2^(2)) .^(20)C_(2) + (3^(2)).^(20)C_(3) +......+(20^(2)).^(20)C_(20)=A(2^(beta)) , then the ordered pair (A, beta) is equal to

(C_(1))/(C_(0))+(2.C_(2))/(C_(1))+(3.C_(3))/(C_(2))+ . . .+(20.C_(20))/(C_(19))=