Each ball can be put in 2 ways (either in one box or the other).
∴ 6 balls can be put in $$2 \times 2 \times .....$$   to six times, i.e., $${2^6}$$ ways. But in two of the ways one box is empty. So, the required number of ways $$ = {2^6} - 2.$$

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 $$= 3! - 2! = 4.$$
$$\therefore $$ The sum of the digits in the unit 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 $$= 3! .$$
$$\therefore $$ The sum of the digits in the thousands place $$ = 6 \times 1 + 6 \times 2 + 6 \times 3 = 36.$$
$$\therefore $$ The required sum $$ = 36 \times 1000 + 24 \times 100 + 24 \times 10 + 24 = 38664.$$

$$\therefore \,\,{T_n} = {\,^n}{C_3}\,;\,{T_{n + 1}} = {\,^{n + 1}}{C_3}$$
As per question,
$$\eqalign{ & {T_{n + 1}} - {T_n} = 21 \cr & \Rightarrow \,{\,^{n + 1}}{C_3}{ - ^n}\,{C_3} = 21 \cr & \frac{{\left( {n + 1} \right)n\left( {n - 1} \right)}}{{3.2.1}} - \frac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{3.2.1}} = 21 \cr & \Rightarrow \,\,n\left( {n - 1} \right)\left( {n + 1 - n + 2} \right) = 126 \cr & \Rightarrow \,\,n\left( {n - 1} \right) = 42 \cr & \Rightarrow \,\,n\left( {n - 1} \right) = 7 \times 6 \cr & \Rightarrow \,\,n = 7. \cr} $$

Total number of ways $$ = \,{\,^3}{C_2} \times {\,^9}{C_2}$$
$$ = 3 \times \frac{{9 \times 8}}{2} = 3 \times 36 = 108$$

$$\eqalign{ & ^{20}{C_1} = {\,^{20}}{C_{19}} = 20 \cr & ^{20}{C_3} = {\,^{20}}{C_{17}} = 20 \times 19 \times 3 \cr & ^{20}{C_5} = {\,^{20}}{C_{15}} = 19 \times 17 \times 16 \times 3 \cr & ^{20}{C_7} = {\,^{20}}{C_{13}} = 19 \times 17 \times 16 \times 15 \cr & ^{20}{C_9} = {\,^{20}}{C_{11}} = 19 \times 17 \times 13 \times 11 \times 5 \times 8 \cr} $$
Clearly G.C.D. $$= 4$$

Clearly, one of the odd digits 1, 3, 5, 7, 9 will be repeated.
The number of selections of the sixth digit $$ = {\,^5}{C_1} = 5.$$
∴ the required number of numbers $$ = \,5 \times \frac{{6!}}{2}.$$

Total number of words that can be formed $$= {10^5}$$
Number of words in which no letter is repeated $$=\, ^{10}{P_5}.$$  So, number of words in which at least one letter is repeated $$ = {10^5} - {\,^{10}}{P_5} = 69760.$$

In $$LUCKNOW,$$    number of letters $$= 7$$  (all distinct), vowels $$\left( {U,O} \right) = 2,$$   consonants $$\left( {L,C,K,N,W} \right) = 5$$
There are 7 letters so 7 places are required. Number of even places are 3 and vowels are 2, so 2 vowels can be placed in $$^3{C_2} \cdot 2! = 6$$   and 5 consonants can be placed in remaining places in $$5!$$ ways.
Hence, number of words $$ = 6 \cdot 5! = 720.$$

$$\eqalign{ & ^n{C_{r + 1}} + {\,^n}{C_{r - 1}} + 2{\,^n}{C_r} = {\,^n}{C_{r - 1}} + {\,^n}{C_r} + {\,^n}{C_r} + {\,^n}{C_{r + 1}} \cr & = {\,^{n + 1}}{C_r} + {\,^{n + 1}}{C_{r + 1}} = {\,^{n + 2}}{C_{r + 1}} \cr} $$

$$\because $$ Each person gets at least one ball.
∴ 3 Persons can have 5 balls in the following systems

Person	I	II	III
No. of balls	1	1	3

or

Person	I	II	III
No. of balls	1	2	2

The number of ways to distribute the balls in first system
$$ = {\,^5}{C_1} \times {\,^4}{C_1} \times {\,^3}{C_3}$$
Also 3, persons having 1, 1 and 3 balls can be arranged $$\frac{{3!}}{{2!}}\,\,{\text{ways}}{\text{.}}$$
∴ No. of ways to distribute 1, 1, 3 balls to the three persons
$$ = {\,^5}{C_1} \times {\,^4}{C_1} \times {\,^3}{C_3} \times \frac{{3!}}{{2!}} = 60$$
Similarly the total no. of ways to distribute 1, 2, 2 balls to the three persons
$$ = {\,^5}{C_1} \times {\,^4}{C_2} \times {\,^2}{C_2} \times \frac{{3!}}{{2!}} = 90$$
∴ The required number of ways = 60 + 90 = 150