$$^n{C_{r - 1}} = 36,{\,^n}{C_r} = 84,{\,^n}{C_{r + 1}} = 126$$
KEY CONCEPT : We know that
$$\eqalign{ & \frac{{^n{C_{r - 1}}}}{{^n{C_r}}} = \frac{r}{{n - r + 1}} \cr & \Rightarrow \,\,\frac{{36}}{{84}} = \frac{r}{{n - r + 1}} \cr & \Rightarrow \,\,\frac{r}{{n - r + 1}} = \frac{3}{7} \cr & \Rightarrow \,\,3n - 10r + 3 = 0\,\,\,\,\,.....\left( 1 \right) \cr & {\text{Also }}\frac{{^n{C_r}}}{{^n{C_{r + 1}}}} = \frac{{r + 1}}{{n - r}} = \frac{{84}}{{126}} = \frac{2}{3} \cr & \Rightarrow \,\,2n - 5r - 3 = 0\,\,\,\,\,.....\left( 2 \right) \cr} $$
Solving (1) and (2), we get
$$n = 9$$  and $$r = 3.$$

The number of committees of 4 gentlemen $$ = {\,^4}{C_4} = 1$$
The number of committees of 3 gentlemen, 1 wife $$ = {\,^4}{C_3} \times {\,^1}{C_1}$$
(∵ after selecting 3 gentlemen only 1 wife is left who can be included)
The number of committees of 2 gentlemen, 2 wives $$ = {\,^4}{C_2} \times {\,^2}{C_2}$$
The number of committees of 1 gentlemen, 3 wives $$ = {\,^4}{C_1} \times {\,^3}{C_3}$$
The number of committees of 4 wives = 1
∴ The required number of committees = 1 + 4 + 6 + 4 + 1 = 16

The number of non-negative integral solutions
= co-efficient of $${x^n}\,{\text{in }}{\left( {{x^0} + {x^1} + {x^2} + .....} \right)^4}$$
= co-efficient of $${x^n}\,{\text{in }}{\left( {1 - x} \right)^{ - 4}}$$
= co-efficient of $${x^n}\,{\text{in }}\left\{ {^3{C_0} + {\,^4}{C_1}x + {\,^5}{C_2}{x^2} + .....} \right\}$$
$$ = {\,^{n + 3}}{C_n} = \frac{{\left( {n + 3} \right)!}}{{n!\,\,3!}} = \frac{{\left( {n + 3} \right)\left( {n + 2} \right)\left( {n + 1} \right)}}{6}.$$

The committee will consist of 4 gentlemen and 1 lady or 3 gentlemen and 2 ladies.
∴ the number of committees $$ = {\,^4}{C_4} \times {\,^6}{C_1}{ + ^4}{C_3} \times {\,^6}{C_2}.$$

Either one boy will be selected or no boy will be selected. Also out of four members one captain is to be selected.
∴ Required number of ways $$ = \left( {^4{C_1} \times {\,^6}{C_3} + {\,^6}{C_4}} \right) \times {\,^4}{C_1} = \left( {80 + 15} \right) \times 4 = 380$$

There are two sets of five parallel lines at equal distances. Clearly, lines like $${l_1},{l_3},{m_1},{m_3}$$   form a square whose diagonal’s length is 2.

∴ the number of required squares
$$ = 3 \times 3$$
$$\left\{ {\because \,\,{\text{choices are }}\left( {{l_1},{l_3}} \right),\left( {{l_2},{l_4}} \right),\left( {{l_3},{l_5}} \right)\,{\text{for one set,e}}{\text{.t}}{\text{.c}}{\text{.}}} \right\}.$$

A selection of four vertices of the polygon gives an interior intersection.

∴ the number of sides $$= n$$
$$\eqalign{ & \Rightarrow \,{\,^n}{C_4} = 70 \cr & \Rightarrow \,\,n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right) = 24 \times 70 = 8 \times 7 \times 6 \times 5 \cr & \therefore \,\,n = 8 \cr} $$
∴ the number of diagonals $$ = {\,^8}{C_2} - 8.$$

$$\eqalign{ & \frac{2}{{9!}} + \frac{2}{{3!7!}} + \frac{1}{{5!5!}} \cr & = \frac{1}{{1!9!}} + \frac{1}{{3!7!}} + \frac{1}{{5!5!}} + \frac{1}{{3!7!}} + \frac{1}{{9!1!}} \cr & = \frac{1}{{10!}}\left\{ {^{10}{C_1} + {\,^{10}}{C_3} + {\,^{10}}{C_5} + {\,^{10}}{C_7} + {\,^{10}}{C_9}} \right\} \cr & = \frac{1}{{10!}}\left( {{2^{10 - 1}}} \right) = \frac{{{2^9}}}{{10!}} = \frac{{{2^a}}}{{b!}}\left( {{\text{given}}} \right) \cr & \Rightarrow a = 9,b = 10 \cr} $$

Total number of required functions = Number of dearrangement of 5 objects
$$ = 5!\left( {\frac{1}{{2!}} - \frac{1}{{3!}} + \frac{1}{{4!}} - \frac{1}{{5!}}} \right) = 44$$

From $$n$$ points number of hexagon is $$^n{C_6} = \frac{{n!}}{{\left\{ {\left( {n - 6} \right)!} \right\}\left\{ {6!} \right\}}}$$
From $$n$$ points number of Octagons is $$^n{C_8} = \frac{{n!}}{{\left\{ {\left( {n - 8} \right)!} \right\}\left\{ {8!} \right\}}}$$
Ratio of number of hexagon to number of octagon is
$$\frac{{\left\{ {\left( {n - 8} \right)!} \right\}\left\{ {8!} \right\}}}{{\left\{ {\left( {n - 6} \right)!} \right\}\left\{ {6!} \right\}}} = \frac{{7 \times 8}}{{\left( {n - 7} \right)\left( {n - 6} \right)}} = \frac{4}{{13}}$$
On solving this quadratic equation we will get $$n = 20$$