Answer: Option A
: A

The total number of 4 digits are 9999-999=9000. The numbers of 4 digits number divisible by 5 are $90\times 20$=1800. Hence required number of ways are 9000-1800 =7200. {Since there are 20 numbers in each hundred (1 to 100) divisible by 5 and from 999 to 9999 there are 90 hundreds, hence the results}.

Answer: Option B
: B

To find the number of times 3 occurs in listing the integer from 1 to 999. (since 3 does notoccur in 1000). Any number between 1 to 999 is a 3 digit number xyzz where the digit x,y,z are any digits from 0 to 9. Now, we first count the numbers in which 3 occurs once only. Since 3 can occur at one place in ${}^3{C}_1$ways, there are ${}^3{C}_1.(9\times 9)$ = ${3.9}^2$such numbers. Again, 3 can occur in exactly two places in ${}^3{C}_1\left(9\right)$ such numbers. Lastly 3 can occur in all the three digits in one such number only 333. $\therefore$The number of times 3 occurs is equal to $1\times (3\times 9^2)+2\times (3\times 9)+3\times 1$=300.

Answer: Option C
: C

The number of selections = coefficient of $x^8$ in $(1+x+x^2+..........+x^8)(1+x+x^2+.........+x^8).(1+x{)}^8$ = coefficient of $x^8$ in $\frac{(1-x^9{)}^2}{(1-x{)}^2}(1+x{)}^8$ = coefficient of $x^8$ in $(1+x{)}^3(1-x{)}^{-2}$ = coefficient of $x^8$ in $({}^8{C}_0+{}^8{C}_{1}x+{}^8{C}_2{x}^2+..........+{}^8{C}_8{x}^8)\times (1+2x+3x^2+4x^3+.........+9x^8+.........)$ = $9.{}^8{C}_0+8.{}^8{C}_1+7.{}^8{C}_2+.............+1.{}^8{C}_8$ = $C_0+2C_{1}x+{}^3{C}_2{x}^2+......{}^9{C}_8{x}^8$ = $(1+x{)}^8+8x(1+x{)}^7$ Putting x = 1, we get $C_0+2C_1+3C_2+........+9C_8$ = $2^8+{8.2}^7$ = $2^7.(1+8)$ = ${10.2}^7$.