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Permutation and Combination Questions Answers

  • 15. From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done

    1. 456
    2. 556
    3. 656
    4. 756
    Answer And Explanation

    Answer: Option D

    Explanation:

    From a group of 7 men and 6 women, five persons are to be selected with at least 3 men.
    So we can have
    (5 men) or (4 men and 1 woman) or (3 men and 2 woman)

    \begin{aligned}
    (^{5}{C}_{5}) + (^{5}{C}_{4} * ^{6}{C}_{1}) + \\
    + (^{5}{C}_{3} * ^{6}{C}_{2}) \\

    = \left[\dfrac{7 \times 6 }{2 \times 1}\right] + \left[\left( \dfrac{7 \times 6 \times 5}{3 \times 2 \times 1} \right) \times 6 \right] + \\ \left[\left( \dfrac{7 \times 6 \times 5}{3 \times 2 \times 1} \right) \times \left( \dfrac{6 \times 5}{2 \times 1} \right) \right] \\
    = 21 + 210 + 525 = 756
    \end{aligned}

  • 16. A bag contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the bag, if at least one black ball is to be included in the draw

    1. 64
    2. 128
    3. 132
    4. 222
    Answer And Explanation

    Answer: Option A

    Explanation:

    From 2 white balls, 3 black balls and 4 red balls, 3 balls are to be selected such that
    at least one black ball should be there.

    Hence we have 3 choices
    All three are black
    Two are black and one is non black
    One is black and two are non black
    Total number of ways
    = 3C3 + (3C2 x 6C1) + (3C1 x 6C2) [because 6 are non black]

    \begin{aligned}
    = 1 + \left[3 \times 6 \right] + \left[3 \times \left(\dfrac{6 \times 5}{2 \times 1}\right) \right]
    = 1 + 18 + 45
    = 64
    \end{aligned}

  • 17. A box contains 4 red, 3 white and 2 blue balls. Three balls are drawn at random. Find out the number of ways of selecting the balls of different colours

    1. 12
    2. 24
    3. 48
    4. 168
    Answer And Explanation

    Answer: Option B

    Explanation:

    This question seems to be a bit typical, isn't, but it is simplest.
    1 red ball can be selected in 4C1 ways
    1 white ball can be selected in 3C1 ways
    1 blue ball can be selected in 2C1 ways

    Total number of ways
    = 4C1 x 3C1 x 2C1
    = 4 x 3 x 2
    = 24

    Please note that we have multiplied the combination results, we use to add when their is OR condition, and we use to multiply when there is AND condition, In this question it is AND as
    1 red AND 1 White AND 1 Blue, so we multiplied.

  • 18. How many words can be formed from the letters of the word "AFTER", so that the vowels never comes together.

    1. 48
    2. 52
    3. 72
    4. 120
    Answer And Explanation

    Answer: Option C

    Explanation:

    We need to find the ways that vowels NEVER come together.
    Vowels are A, E
    Let the word be FTR(AE) having 4 words.
    Total ways = 4! = 24
    Vowels can have total ways 2! = 2
    Number of ways having vowel together = 48
    Total number of words using all letter = 5! = 120

    Number of words having vowels never together = 120-48
    = 72

  • 19. In a Cricket cup total 153 matches were played and every two teams played exactly one match with each other. So what were the total number of teams participating in Cricket Cup ?

    1. 15
    2. 16
    3. 17
    4. 18
    5. 19
    Answer And Explanation

    Answer: Option D

    Explanation:

    Lets suppose there were x number os team in cricket cup then from the given statement we can calculate :
    \begin{aligned}
    => ^n C _2 = 153 \\
    \text{ it given n = 18 and n = -17} \\
    \end{aligned}
    As answer cannot be negative to 18 is the answer