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- 1. In how many different number of ways 4 boys and 2 girls can sit on a bench?

A.) 620

B.) 640

C.) 720

D.) None of these

Answer: Option ''

Answer: Option 'C'

^{n}p_{n} = n!

^{6}p_{6}

= 6 × 5 × 4 × 3 × 2 × 1 = 720

- 2. In how many different number of ways 5 men and 2 women can sit on a shopa which can accommodate persons?

A.) 230

B.) 203

C.) 220

D.) 210

Answer: Option 'D'

Answer: Option 'D'

^{7}p_{3} = 7 × 6 × 5 = 210

- 3. In how many different number of ways 4 boys and 3 girls can sit on a bench such that girls always sit together.

A.) 720

B.) 5040

C.) 4320

D.) None of these

Answer: Option 'A'

Answer: Option 'A'

- 4. In how many different ways can the letters of the word "CLAIM" be rearrangement?

A.) 120

B.) 125

C.) 130

D.) None of these

Answer: Option 'A'

Answer: Option 'A'

The total number of arrangements is

^{5}P_{5} = 5! = 120

- 5. If the letters of the word PLACE are arranged taken all at a time, find how many do not start with AE.

A.) 142

B.) 141

C.) 114

D.) None of these

Answer: Option 'C'

Answer: Option 'C'

Total no'of arrangements ^{5}P_{5} = 5! = 120

no'of arrangements start with AE = 1 × 6 = 6

no'of arrangements which do not start with AE = 120 - 6 = 114.

- 6. How many arrangements of the letters of the word BEGIN can be made, without changing the place of the vowels in the word?

A.) 7 ways

B.) 6 ways

C.) 5 ways

D.) 2 ways

Answer: Option ''

Answer: Option 'B'

E,I fixed. Consonants can be arrangements in ^{3}P_{3} = 3! = 6 ways

- 7. If all the numbers 2, 3, 4, 5, 6, 7, 8 are arranged, find the number of arrangements in which 2, 3, 4, are together?

A.) 720

B.) 620

C.) 700

D.) None of these

Answer: Option 'A'

Answer: Option 'A'

If (2 3 4) is one.

we must arrange (2 3 4), 5, 6, 7, 8 in

^{5}P_{5} = 5! = 120 ways

2, 3, 4 can be arranged in ^{3}P_{3} = 3! = 6

120 × 6 = 720.

- 8. Find
^{10}P_{6}

A.) 150200

B.) 151200

C.) 152200

D.) None of these

Answer: Option ''

Answer: Option 'B'

^{10}P_{6} = 10!/4! = 10 × 9 × 8 × 7 × 6 × 5

= 151200.

- 9. Find
^{9}P_{3}

A.) 414

B.) 514

C.) 504

D.) None of these

Answer: Option 'B'

^{9}P_{3} = 9!/6! = 9 × 8 × 7

= 504.

- 10. Find
^{7}P_{7}

A.) 4440

B.) 5040

C.) 5045

D.) None of these

Answer: Option 'B'

^{7}P_{7} = 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040

**To whom this Permutations Combinations Question and Answers section is beneficial?**

Students can learn and improve on their skillset for using Permutations Combinations effectively and can also prepare for competitive examinations like...

- All I.B.P.S and Public Sector Bank Competitive Exam
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