Let $$y = p + 1, z = q + 2.$$     Then $$x \geqslant 0,p \geqslant 0,q \geqslant 0$$    and $$x + y + z = 15$$    implies $$x + p + q = 12.$$
∴ the required number of values of $$(x, y, z)$$  and hence of $$(x, p, q)$$
= the number of non-negative integral solutions of $$(x + p + q = 12)$$
= co-efficient of $${x^{12}}\,{\text{in }}{\left( {{x^0} + {x^1} + {x^2} + .....} \right)^3}$$
= co-efficient of $${x^{12}}\,{\text{in }}{\left( {1 - x} \right)^{ - 3}}$$
= co-efficient of $${x^{12}}\,{\text{in }}\left\{ {^2{C_0} + {\,^3}{C_1}x + {\,^4}{C_2}x + .....} \right\}$$
$$ = {\,^{14}}{C_{12}} = \frac{{14!}}{{12!\,\,2!}} = \frac{{14 \times 13}}{2} = 91.$$

$$\eqalign{ & ^{x + 2}{P_{x + 2}} = a \cr & \Rightarrow a = \left( {x + 2} \right)! \cr & ^x{P_{11}} = b \cr & \Rightarrow b = \frac{{x!}}{{\left( {x - 11} \right)!}}s{\text{ and}}{{\text{ }}^{x - 11}}{P_{x - 11}} = c \cr & \Rightarrow c = \left( {x - 11} \right)!;a = 182\,bc \cr & \therefore \left( {x + 2} \right)! = 182\frac{{x!}}{{\left( {x - 11} \right)!}}\left( {x - 11} \right)! \cr & \Rightarrow \left( {x + 2} \right)\left( {x + 1} \right) = 182 = 14 \times 13 \cr & \therefore x + 1 = 13 \cr & \therefore x = 12 \cr} $$

We have, $$A = \left\{ {2,3,5,7,11,13,17,19,23,29} \right\}.$$        A contains 10 elements. So numerator and denominator each can be chosen in 10 ways.
So no. of rational numbers
$$= 10 \times 10 - 10 + 1 = 91$$
(Out of these selections, 10 numbers are simply 1)

$$1800 = {2^3} \times {3^2} \times {5^2}$$
∴ the required number of proper divisors
= total number of selections of at least one 2 and one 5 from 2, 2, 2, 3, 3, 5, 5
$$ = 3 \times \left( {2 + 1} \right) \times 2 = 18.$$

As $$0 < x < 1,$$   we have $$p < q.$$
The number of rational numbers $$= 5 + 4 + 3 + 2 + 1 = 15.$$
When $$p, q$$  have a common factor, we get some rational numbers which are not different from those already counted. There are 4 such numbers:
$$\frac{2}{4},\frac{2}{6},\frac{3}{6},\frac{4}{6}.$$
∴ the required number of rational numbers $$ = 15 - 4 = 11.$$

Required number of numbers
$$ = 3 \times 5 \times 5 \times 5 = 375$$

If $$r$$ questions are solved by each student then the number of possible selections of questions is $$^{20}{C_r}.$$
∴ the number of students $$= {^{20}{C_r}}.$$
($$\because $$ each student has solved different combinations of questions)
∴ the maximum number of students = maximum value of $$^{20}{C_r} = {\,^{20}}\,{C_{10}},$$   because $$^{20}\,{C_{10}}$$  is the largest among $$^{20}{C_0},{\,^{20}}{C_1},.....{,^{20}}{C_{20}} - $$     being the middle one.

$$\because $$ Card numbered 1 is always placed in envelope numbered 2, we can consider two cases.
Case I : Card numbered 2 is placed in envelope numbered 1.
Then it is dearrangement of 4 objects, which can be done in $$4!\left( {1 - \frac{1}{{1!}} + \frac{1}{{2!}} - \frac{1}{{3!}} + \frac{1}{{4!}}} \right) = 9\,\,{\text{ways}}$$
Case II : Card numbered 2 is not placed in envelope numbered 1.
Then it is dearrangement of 5 objects, which can be done in $$5!\left( {1 - \frac{1}{{1!}} + \frac{1}{{2!}} - \frac{1}{{3!}} + \frac{1}{{4!}} - \frac{1}{{5!}}} \right) = 44\,\,{\text{ways}}$$
∴ Total ways = 44 + 9 = 53

No. of words starting with $$A$$ are $$4 ! = 24$$
No. of words starting with $$H$$ are $$4 ! = 24$$
No. of words starting with $$L$$ are $$4 ! = 24$$
These account for 72 words
Next word is $$RAHLU$$   and the $$74^{th}$$  word $$RAHUL.$$

Required number of ways $$ = {\text{coeff}}{\text{. of }}{x^{16}}{\text{ in}}{\left( {{x^3} + {x^4} + {x^5} + {x^6} + {x^7}} \right)^4}$$