Permutations and Combinations Practice Problems: Level 02

Directions for questions 1-10: Solve the following combination and permutation questions as per the best of your abilities. You may have to apply combination and permutation formula to answer some of these questions.
  1. Find the sum of all the 4 digit numbers that can be formed with the digits 3, 4, 5 and 6
    1. 119988
    2. 11988
    3. 191988
    4. None of these

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    Answer : Option 1
    No. of Digits = 4 All are distinct; They can be arranged in 4! = 24 ways
    Each of the digits 3, 4, 5 and 6 occur at unit place = 3! Ways = 6 ways.
    Thus there will be 6 numbers ending with 3, 4, 5 and 6 each. So the sum of the digits at unit's place = 6(3 + 4 + 5 + 6) =108
    The sum of numbers = 108 × 103 + 108 × 102 + 108 × 101 + 108 × 100 = 119988
  2. Find the sum of all the 4 digit numbers that can be formed with the digits 3, 4, 4 and 2.
    1. 43339
    2. 43999
    3. 43329
    4. None of these

    Answer : Option 3
    Here each of the digits 2 and 3 will occur at unit, tens, hundred and thousand place ( 3P3/2!) = 3 times. Digit 4 will occur at each place = 6 times;
    ∴ Sum of digits at unit, tens, hundred and thousand place = 3 × 3 + 6 × 4 + 3 × 2 =39. Sum of numbers formed =
    = 39 × 103 + 39× 102 + 39 × 101 + 39 × 100 = 43329
  3. The number of straight lines that can be drawn out of 12 points of which 8 are collinear is
    1. 39
    2. 29
    3. 49
    4. 59

    Answer : Option 1
    The required number of lines= 12C28C2 + 1 = 1 + 66 – 28 = 39
  4. A box contains three white balls, four black balls and three red balls. The number of ways in which three balls can be drawn from the box so that at least one of the balls is black is
    1. 50
    2. 100
    3. 150
    4. 200

    Answer : Option 2
    The required number of ways
    (a) 1 black and 2 others = 4C1.6C2 = 4 × 15 = 60
    (b) 2 black and 1 other = 4C2.6C1 = 6 × 6 = 36
    (c) All the three black = 4C3 = 4
    Total =60 + 36 + 4 = 100
  5. In a cricket championship, there are 21 matches. If each team plays one match with every other team, the number of teams is
    1. 7
    2. 9
    3. 10
    4. None of these

    Answer : Option 1
    Let n be the number of teams.
    nC2 = 21
    (n(n-1)/2) = 21
    ⇒ n(n-1) = 42 ∴
    ⇒ n = 7
  1. The number of four–digit telephone numbers having at least one of their digits repeated is
    1. 9,000
    2. 1,00,00
    3. 3,240
    4. 4960

    Answer : Option 4
    The number of four–digit telephone numbers which can be formed using the digits of 0, 1, 2,...., 9 is 104.
    The number of four digit telephone numbers which have none of their digits repeated is 10P4 = 10 ×9 ×8 ×7 = 5040
    Hence the required number =104- 5040 = 4960
  2. In how many ways can you rearrange the word JUMBLE such that the rearranged word starts with a vowel?
    1. 120
    2. 240
    3. 360
    4. 60

    Answer : Option 2
    JUMBLE is a six-lettered word. Since the rearranged word has to start with a vowel, the first letter can be either U or E. The balance 5 letters can be arranged in 5P5 or 5! ways. Total number of words = 2 × 5! = 240.
  3. In an examination, a candidate is required to pass all five different subjects. The number of ways he can fail is:
    1. 32
    2. 31
    3. 30
    4. 29

    Answer : Option 2
    The candidate will fail if he fails either in 1 or 2 or 3 or 4 or 5 subjects, ∴ Required number of ways 5C1 + 5C2 + 5C3 + 5C4 + 5C5 = 31
  4. The sum of the divisors of 26.34.57 is
    1. 26.38.54.73
    2. 27.35.58-2.3.5
    3. 26.38.54.73-1
    4. None of these

    Answer : Option 4
    Any divisor of 26.34.57. is of the form 2a.3b.5c. where 0≤ a ≤ 6, 0 ≤ b ≤ 4 & 0 ≤ c ≤ 7.
    Thus the sum of the divisors of 26.34.57.
    is (1 + 2+…..+25 + 26)(1 + 3 + …. + 34)(1 + 5 +....+ 57) =
    ((27-1)/(2-1))((35-1)/(3-1))((58-1)/(5-1))
    =((27-1)(35-1)(58-1)/2.4)
  5. Nine chairs are numbered 1 to 9. Three women and four men wish to occupy one chair each. First the women chose the chairs from amongst the chair marked 1 to 5; and then the men select the chairs from amongst the remaining. The number of possible arrangements is
    1. 5C3 × 4C2
    2. 5C2 × 4P3
    3. 5C3 × 6C4
    4. None of these

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    Answer : Option 3
    Women can select 3 chairs from chairs numbered 1 to 5 in 5C3 ways and remaining 6 chairs can be selected by 4 men in 6C4 ways. Hence the required number of ways = 5C3 × 6C4