Permutation and Combination

21. How many positive integers less than 100,000 and divisible by 125 can be formed using the digits 0, 1, 2, 5 and 8, if repetition is allowed?

  • 90
  • 300
  • 99
  • 129
Show Explanation
Correct Answer: 99
A number is divisible by 125 if and only if the number formed by its last three digits is divisible by 125. Possible last three digits: 000, 125, 250, 500. Three-digit numbers = 3 Four-digit numbers = 4 × 4 = 16 Five-digit numbers = 4 × 4 × 5 = 80 Total = 3 + 16 + 80 = 99.

22. Two red pencils, three black pencils and two white pencils are to be arranged in a row such that: (a) no two adjacent pencils are of the same colour and (b) the pencils at the two ends of the row are of same colour. In how many ways can the pencils be arranged?

  • 12
  • 8
  • 9
  • 10
Show Explanation
Correct Answer: 10
The possible arrangements satisfying both conditions are: RBWBWBR, WBRBRBW, BWBRWRB, BRBWRWB, BWRBWRB, BWRRBRWB, BRWBRWB, BRWBWRB, BWRWBRB and BRWBBWB. Total = 10.

23. Mangalam has forgotten his friend’s 8-digit telephone number but remembers the following: (a) The first 3 digits are either 270 or 279. (b) The digit 0 occurs exactly three times and the digit 9 occurs exactly once. (c) The number is an even number. If Mangalam were to use a trial and error method to reach his friend, what is the maximum number of trials he has to make to be sure to succeed?

  • 832
  • 1664
  • 1280
  • 2000
Show Explanation
Correct Answer: 1664
The last digit must be 0, 2, 4, 6 or 8. Beginning with 270 and ending with 0 → 768 ways Beginning with 270 and ending with 2,4,6,8 → 384 ways Beginning with 279 and ending with 0 → 384 ways Beginning with 279 and ending with 2,4,6,8 → 128 ways Total = 768 + 384 + 384 + 128 = 1664.

24. There is an unlimited supply of identical red, blue and green coloured balls. In how many ways can 12 balls be selected from the supply?

  • 66
  • 78
  • 91
  • 105
Show Explanation
Correct Answer: 91
Let x₁ + x₂ + x₃ = 12. Number of non-negative integral solutions = (12 + 3 − 1)C(3 − 1) = 14C2 = 91.

25. How many numbers between 4000 and 6000 can be formed using the digits 1 to 6, when any digit can occur any number of times?

  • 864
  • 432
  • 638
  • 126
Show Explanation
Correct Answer: 432
First digit can be 4 or 5 → 2 ways. Remaining digits → 6 × 6 × 6 ways. Total = 2 × 6³ = 432.

26. In how many ways can a group of 15 persons be arranged around two circular tables consisting of 7 and 8 chairs?

  • 15!6!7!/8!
  • 15!6!/8!
  • 15!/7!
  • 15
Show Explanation
Correct Answer: 15!6!/8!
Select 7 persons out of 15 → 15C7 Arrange them around table → 6! Arrange remaining 8 → 7! Total = 15C7 × 6! × 7! = 15!6!/8!.

27. The number of positive integral solutions of the equation a + b + c = 15 is _______.

  • 76
  • 105
  • 91
  • 86
Show Explanation
Correct Answer: 91
Number of positive integral solutions of a + b + c = 15 = (15 − 1)C(3 − 1) = 14C2 = 91.

28. The number of non-negative integral solutions of the equation p + q + r + s = 30 is _______.

  • 4598
  • 5324
  • 5546
  • 5456
Show Explanation
Correct Answer: 5456
Number of non-negative integral solutions of p + q + r + s = 30 = (30 + 4 − 1)C(4 − 1) = 33C3 = (33 × 32 × 31) / (3 × 2 × 1) = 5456.

29. What is the number of non-negative integer solutions of A + B + C + D ≤ 15?

  • 18C3
  • 19C3
  • 18C4
  • 19C4
Show Explanation
Correct Answer: 19C4
Given A + B + C + D ≤ 15 Let A + B + C + D = 15 − E ⇒ A + B + C + D + E = 15 Now A, B, C, D, E are non-negative integers. Number of solutions = (15 + 5 − 1)C(5 − 1) = 19C4.

30. In how many ways can five letters be posted into 3 post boxes such that at least one letter is posted in each box?

  • 120
  • 150
  • 180
  • 210
Show Explanation
Correct Answer: 150
Let B1, B2 and B3 be the three boxes. Total number of ways = 3(20) + 3(30) = 150.