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:

If    $${{{}^{n + 2}C{}_6} \over {{}^{n - 2}{P_2}}}$$ = 11, then n satisfies the equation :

The sum $$\sum\limits_{r = 1}^{10} {\left( {{r^2} + 1} \right) \times \left( {r!} \right)} $$ is equal to :

If the four letter words (need not be meaningful ) are to be formed using the letters from the word “MEDITERRANEAN” such that the first letter is R and the fourth letter is E, then the total number of all such words is :