Permutation and Combination Problems and Solutions

• 8. Evaluate combination
  \begin{aligned}
  ^{100}{C}_{97} = \frac{100!}{(97)!(3)!} \\
  \end{aligned}

  1. 161700
  2. 151700
  3. 141700
  4. 131700

  Answer :

  Option A

  Explanation:

  \begin{aligned}
  ^{n}{C}_r = \frac{n!}{(r)!(n-r)!} \\
  ^{100}{C}_{97} = \frac{100!}{(97)!(3)!} \\
  = \frac{100*99*98*97!}{(97)!(3)!} \\
  = \frac{100*99*98}{3*2*1} \\
  = \frac{100*99*98}{3*2*1} \\
  = 161700
  \end{aligned}

• 9. 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 :

  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.

• 10. In how many words can be formed by using all letters of the word BHOPAL

  1. 420
  2. 520
  3. 620
  4. 720

  Answer :

  Option D

  Explanation:

  Required number
  \begin{aligned}
  = 6! \\
  = 6*5*4*3*2*1 \\
  = 720
  \end{aligned}
  

• 11. In how many ways can the letters of the CHEATER be arranged

  1. 20160
  2. 2520
  3. 360
  4. 80

  Answer :

  Option B

  Explanation:

  As we can see the letter "E" is twice in given word, so Required Number
  \begin{aligned}
  = \frac{7!}{2!} \\
  = \frac{7*6*5*4*3*2!}{2!} \\
  = 2520
  \end{aligned}

• 12. In how many way the letter of the word "APPLE" can be arranged

  1. 20
  2. 40
  3. 60
  4. 80

  Answer :

  Option C

  Explanation:

  Friends the main point to note in this question is letter "P" is written twice in the word.
  Easy way to solve this type of permutation question is as,
  
  So word APPLE contains 1A, 2P, 1L and 1E
  Required number =
  \begin{aligned}
  = \frac{5!}{1!*2!*1!*1!} \\
  = \frac{5*4*3*2!}{2!} \\
  = 60
  \end{aligned}
  

• 13. Evaluate combination
  \begin{aligned}
  ^{100}{C}_{100}
  \end{aligned}

  1. 10000
  2. 1000
  3. 10
  4. 1

  Answer :

  Option D

  Explanation:

  \begin{aligned}
  ^{n}{C}_{n} = 1 \\
  ^{100}{C}_{100} = 1
  \end{aligned}

• 14. 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 :

  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}