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Question 1. The number of numbers of 4 digits which are not divisible by 5 are
  1.    7200
  2.    3600
  3.    14400
  4.    1800
Answer: Option A
: A

The total number of 4 digits are 9999-999=9000. The numbers of 4 digits number divisible by 5 are 90×20=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}.

Question 2. The number of times the digit 3 will be written when listing the integers from 1 to 1000 is
  1.    269
  2.    300
  3.    271
  4.    302
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 3C1ways, there are 3C1.(9×9) = 3.92such numbers. Again, 3 can occur in exactly two places in 3C1(9) such numbers. Lastly 3 can occur in all the three digits in one such number only 333. The number of times 3 occurs is equal to 1×(3×92)+2×(3×9)+3×1=300.

Question 3. Number of ways of selection of 8 letters from 24 letters of which 8 are a, 8 are b and the rest unlike, is given by
  1.    27
  2.    8.28
  3.    10.27
  4.    8.27
Answer: Option C
: C

The number of selections = coefficient of x8 in (1+x+x2+..........+x8)(1+x+x2+.........+x8).(1+x)8 = coefficient of x8 in (1x9)2(1x)2(1+x)8 = coefficient of x8 in (1+x)3(1x)2 = coefficient of x8 in (8C0+8C1x+8C2x2+..........+8C8x8)×(1+2x+3x2+4x3+.........+9x8+.........) = 9.8C0+8.8C1+7.8C2+.............+1.8C8 = C0+2C1x+3C2x2+......9C8x8 = (1+x)8+8x(1+x)7 Putting x = 1, we get C0+2C1+3C2+........+9C8 = 28+8.27 = 27.(1+8) = 10.27.