$$\eqalign{ & ^{n + 1}{C_3} - {\,^n}{C_3} = 21 \cr & \because {\,^n}{C_r} + {\,^n}{C_{r - 1}} = {\,^{n + 1}}{C_r} \cr & \Rightarrow {\,^n}{C_2} = 21 \cr & \Rightarrow n = 7 \cr} $$

Let the sides of the game be $$A$$ and $$B.$$ Given 5 married couples, i.e., 5 husbands and 5 wives. Now, 2 husbands for two sides $$A$$ and $$B$$ can be selected out of 5 $$ = {\,^5}{C_2} = 10{\text{ ways}}{\text{.}}$$
After choosing the two husbands their wives are to be excluded (since no husband and wife play in the same game). So we have to choose 2 wives out of remaining $$5 - 2 = 3$$   wives i.e., $$ = {\,^3}{C_2} = 3{\text{ ways}}{\text{.}}$$
Again two wives can interchange their sides $$A$$ and $$B$$ in $$2! = 2{\text{ ways}}{\text{.}}$$
By the principle of multiplication, the required number of ways $$ = 10 \times 3 \times 2 = 60$$

The natural numbers are 1, 2, 3, 4.

Clearly, in one diagonal we have to place 1, 4 and in the other 2, 3.
The number of ways in (i) $$ = 2!\, \times 2! = 4.$$
The number of ways in (ii) $$ = 2!\, \times 2! = 4.$$
∴ the total number of ways = 8.

The straight line $${l_1},{l_2},{l_3}$$  are parallel and lie in the same plane.
Total number of points $$= m + n + k$$
Total no. of triangles formed with vertices $$ = {\,^{m + n + k}}{C_3}$$
By joining three given points on the same line we don’t obtain a triangle.
Therefore, the max. number of triangles $$ = {\,^{m + n + k}}{C_3} - {\,^m}{C_3} - {\,^n}{C_3} - {\,^k}{C_3}$$

The numbers will be made by 0, 2, 4, 6 or 0, 4, 6, 8
∴ the required number of numbers $$ = \left( {^4{P_4} - {\,^3}{P_3}} \right) + \left( {^4{P_4} - {\,^3}{P_3}} \right).$$

The number of numbers with 0 in the units place $$= 3! = 6.$$
The number of numbers with 1 or 2 or 3 in the units place $$= 3! - 2! = 4.$$
∴ the sum of the digits in the units place $$ = 6 \times 0 + 4 \times 1 + 4 \times 2 + 4 \times 3 = 24.$$
Similarly for the tens and the hundreds places.
The number of numbers with 1 or 2 or 3 in the thousands place $$= 3!.$$
∴ the sum of the digits in the thousands place $$ = 6 \times 1 + 6 \times 2 + 6 \times 3 = 36.$$
∴ the required sum $$ = 36 \times 1000 + 24 \times 100 + 24 \times 10 + 24.$$