Permutation and Combination Problems and Solutions

• 1. Evaluate permutation equation
  \begin{aligned} ^{75}{P}_2\end{aligned}

  1. 5200
  2. 5300
  3. 5450
  4. 5550

  Answer :

  Option D

  Explanation:

  \begin{aligned}
  ^n{P}_r = \frac{n!}{(n-r)!} \\
  ^{75}{P}_2 = \frac{75!}{(75-2)!} \\
  = \frac{75*74*73!}{(73)!} \\
  = 5550
  
  
  \end{aligned}

• 2. In how many ways can the letters of the word "CORPORATION" be arranged so that vowels always come together.

  1. 5760
  2. 50400
  3. 2880
  4. None of above

  Answer :

  Option B

  Explanation:

  Vowels in the word "CORPORATION" are O,O,A,I,O
  Lets make it as CRPRTN(OOAIO)
  
  This has 7 lettes, where R is twice so value = 7!/2!
  = 2520
  
  Vowel O is 3 times, so vowels can be arranged = 5!/3!
  
  = 20
  
  Total number of words = 2520 * 20 = 50400

• 3. How many words can be formed by using all letters of TIHAR

  1. 100
  2. 120
  3. 140
  4. 160

  Answer :

  Option B

  Explanation:

  First thing to understand in this question is that it is a permutation question.
  Total number of words = 5
  Required number =
  \begin{aligned}
  ^5{P}_5 = 5! \\
  = 5*4*3*2*1 = 120
  \end{aligned}

• 4. Evaluate permutation equation
  \begin{aligned} ^{59}{P}_3 \end{aligned}

  1. 195052
  2. 195053
  3. 195054
  4. 185054

  Answer :

  Option C

  Explanation:

  \begin{aligned}
  ^n{P}_r = \frac{n!}{(n-r)!} \\
  ^{59}{P}_3 = \frac{59!}{(56)!} \\
  = \frac{59 * 58 * 57 * 56!}{(56)!} \\
  = 195054
  \end{aligned}

• 5. Evaluate permutation
  \begin{aligned}
  ^5{P}_5
  \end{aligned}

  1. 120
  2. 110
  3. 98
  4. 24

  Answer :

  Option A

  Explanation:

  \begin{aligned}
  ^n{P}_n = n! \\
  ^5{P}_5 = 5*4*3*2*1 \\
  = 120
  
  \end{aligned}

• 6. In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there

  1. 109
  2. 128
  3. 138
  4. 209

  Answer :

  Option D

  Explanation:

  In a group of 6 boys and 4 girls, four children are to be selected such that
  at least one boy should be there.
  So we can have
  
  (four boys) or (three boys and one girl) or (two boys and two girls) or (one boy and three gils)
  
  This combination question can be solved as
  
  \begin{aligned}
  (^{6}{C}_{4}) + (^{6}{C}_{3} * ^{4}{C}_{1}) + \\
  + (^{6}{C}_{2} * ^{4}{C}_{2}) + (^{6}{C}_{1} * ^{4}{C}_{3}) \\
  
  = \left[\dfrac{6 \times 5 }{2 \times 1}\right] + \left[\left(\dfrac{6 \times 5 \times 4 }{3 \times 2 \times 1}\right) \times 4\right] + \\\left[\left(\dfrac{6 \times 5 }{2 \times 1}\right)\left(\dfrac{4 \times 3 }{2 \times 1}\right)\right] + \left[6 \times 4 \right] \\
  = 15 + 80 + 90 + 24\\
  = 209
  \end{aligned}

• 7. In how many way the letter of the word "RUMOUR" can be arranged

  1. 2520
  2. 480
  3. 360
  4. 180
  5. is more today than ever before
  6. was always there
  7. is no longer there
  8. will always be there

  Answer :

  Option E

  Explanation:

  In above word, there are 2 "R" and 2 "U"
  So Required number will be
  
  \begin{aligned}
  = \frac{6!}{2!*2!} \\
  = \frac{6*5*4*3*2*1}{4} \\
  = 180
  \end{aligned}