Number of white balls $$= 10$$
Number of green balls $$= 9$$
and Number of black balls $$= 7$$
∴ Required probability $$= (10 + 1) (9 + 1)(7 + 1) - 1 = 11 \cdot 10 \cdot 8 - 1 = 879$$
[ $$\because $$ The total number of ways of selecting one or more items from $$p$$ identical items of one kind, $$q$$ identical items of second kind; $$r$$ identical items of third kind is $$(p + 1)(q + 1 )(r + 1) - 1 ] $$

Consider the equation $$a + b = 10$$   number of solutions of this equation is $$^{10 + 2 - 1}{C_{2 - 1}} = 11.$$
Next equation is $$a + b + c + d = 21$$    hence $$c + d = 11$$   and number of solutions of this equation is 12.
Similarly for third equation $$a + b + c + d + e + f = 33$$      or $$e + f = 12$$   or number of solutions is 13.
Similarly for last equation $$a + b + c + d + ..... + x + y + z = 208,$$       or $$y + z = 22$$   or number of solution is 23.
Required number of ways is $$11 \times 12 \times 13 \times ..... \times 21 \times 22 \times 23 = \frac{{23!}}{{10!}} = {\,^{23}}{P_{13}}.$$

$$\eqalign{ & {\text{Here, }}E = \left( {2n + 1} \right)\left( {2n + 3} \right)\left( {2n + 5} \right).....\left( {4n - 3} \right)\left( {4n - 1} \right) \cr & {\text{or, }}E = \frac{{\left( {2n + 1} \right) \cdot \left( {2n + 2} \right) \cdot \left( {2n + 3} \right) \cdot \left( {2n + 4} \right).....\left( {4n - 1} \right)\left( {4n} \right)}}{{\left( {2n + 2} \right)\left( {2n + 4} \right).....\left( {4n} \right)}} \cdot \frac{{\left( {2n} \right)!}}{{\left( {2n} \right)!}} \cr & E = \frac{{\left( {4n} \right)! \cdot n!}}{{\left( {2n} \right)!{2^n} \cdot \left( {2n} \right)!}} \cr & \Rightarrow {2^n}E = \frac{{\left( {4n} \right)!}}{{\left( {2n} \right)!\left( {2n} \right)!}} \cdot n! = {\,^{4n}}{C_{2n}} \cdot n! \cr & \Rightarrow {2^n}E{\text{ is divisible by}}{{\text{ }}^{4n}}{C_{2n}}. \cr} $$

L	O	Y
2	2	4
$$ \geqslant 1$$	$$ \geqslant 1$$	$$2 \leqslant $$

⟹

L	O	Y
1	1	2
1	2	1
2	1	1
2	2	0

Required number of ways
$$\eqalign{ & = {\,^2}{C_1} \times {\,^2}{C_1} \times {\,^4}{C_2} \times {\,^2}{C_1} \times {\,^2}{C_2} \times {\,^4}{C_1} + {\,^2}{C_2} \times {\,^2}{C_1} \times {\,^4}{C_1} + {\,^2}{C_2} \times {\,^2}{C_2} \times {\,^4}{C_0} \cr & = 2 \times 2 \times \frac{{4 \times 3}}{2} + 2 \times 1 \times 4 + 1 \times 2 \times 4 + 1 \times 1 \times 1 \cr & = 24 + 8 + 8 + 1 = 41. \cr} $$

$$\eqalign{ & 240 = {2^4} \times 3 \times 5 \cr & 4n + 2 = 2\left( {2n + 1} \right) = 2 \times \,{\text{odd}} \cr} $$
∴ the required number of divisors
= the number of selections of one 2 from four $$2’s,$$  any number of $$3’s$$  from one 3 and any number of $$5’s$$  from one 5.
$$ = 1 \times 2 \times 2 = 4.$$

$$A$$ = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29}. A rational number is made by taking any two in any order.
∴ the required number of rational numbers $$ = {\,^{10}}{P_2} + 1$$   (including 1).

The number of circles $$ = \left( {^{10}{C_3} - {\,^4}{C_3}} \right) + 1.$$

Here in this case condition is similar to formation of necklace i.e.,
$$\left( {n,k} \right) = \frac{1}{n}\sum\limits_{i = 1}^n {{k^{\gcd \left( {n,i} \right)}}} $$
We can use this formula or from the table (you shouldn’t memorize it) required number of ways is 70.

If $$N = {p_1}^{{\alpha _1}}{p_2}^{{\alpha _2}}$$   then the sum of the divisors of $$N$$ is
$$\left( {\frac{{{p_1}^{{\alpha _1} + 1} - 1}}{{{p_1} - 1}}} \right)\left( {\frac{{{p_2}^{{\alpha _2} + 1} - 1}}{{{p_2} - 1}}} \right)$$

Required number of ways
$$ = {\text{coefficient of }}{x^8}{\text{ in}}{\left( {x + {x^2} + .....{x^6}} \right)^3}$$
[∵ each box can receive minimum 1 and maximum 6 balls]
$$\eqalign{ & = {\text{coeff}}{\text{. of }}{x^8}{\text{ in }}{x^2}{\left( {1 + x + {x^2} + ..... +{x^5}} \right)^3} \cr & = {\text{coeff}}{\text{. of }}{x^5}{\text{ in}}{\left( {\frac{{1 - {x^6}}}{{1 - x}}} \right)^3} \cr & = {\text{coeff}}{\text{. of }}{x^5}{\text{ in}}{\left( {1 - x} \right)^{ - 3}} \cr & = {\text{coeff}}{\text{. of }}{x^5}{\text{ in}}\left( {1 + {\,^3}{C_1}x + {\,^4}{C_2}x^2 + .....} \right) \cr & = {\,^7}{C_5} \cr & = 21 \cr} $$