If for a positive integer n, the quadratic equation

$$x\left( {x + 1} \right) + \left( {x + 1} \right)\left( {x + 2} \right)$$$$ + .... + \left( {x + \overline {n - 1} } \right)\left( {x + n} \right)$$$$ = 10n$$

has two consecutive integral solutions, then n is equal to :

A man X has 7 friends, 4 of them are ladies and 3 are men. His wife Y also has 7 friends, 3 of them are ladies and 4 are men. Assume X and Y have no common friends. Then the total number of ways in which X and Y together can throw a party inviting 3 ladies and 3 men, so that 3 friends of each of X and Y are in this party, is:

The value of $$\left( {{}^{21}{C_1} - {}^{10}{C_1}} \right) + \left( {{}^{21}{C_2} - {}^{10}{C_2}} \right) + \left( {{}^{21}{C_3} - {}^{10}{C_3}} \right)$$
$$\left( {{}^{21}{C_4} - {}^{10}{C_4}} \right)$$$$ + .... + \left( {{}^{21}{C_{10}} - {}^{10}{C_{10}}} \right)$$ is

Let $$a$$, b, c $$ \in R$$. If $$f$$(x) = ax² + bx + c is such that
$$a$$ + b + c = 3 and $$f$$(x + y) = $$f$$(x) + $$f$$(y) + xy, $$\forall x,y \in R,$$

then $$\sum\limits_{n = 1}^{10} {f(n)} $$ is equal to