BANK & INSURANCE (PERMUTATION AND COMBINATION) PART 1

Total Questions: 30

1. In how many different ways can the letters of the word “PHONE” be arranged so that the vowels may occupy only odd positions?

Correct Answer: (e) None of these
Solution:

Number of ways of arranging the vowels in odd positions = ³P₂ = 3 × 2 = 6
Number of ways of arranging the remaining letters = 3! = 6
Required number of ways = 6 × 6 = 36

2. In how many ways can a group of 4 men and 3 women can be made out of a total of 6 men and 5 women?

Correct Answer: (a) 150  
Solution:

Required ways = ⁶C₄ and ⁵C₃
⇒ [(6 × 5 × 4 × 3)/(1 × 2 × 3 × 4)] × [(5 × 4 × 3)/(1 × 2 × 3)]
⇒ 15 × 10 = 150

3. In how many ways can the person sent letters to 10 different persons, if the letter distributed by three different postmen?

Correct Answer: (d) 310  
Solution:

Total ways = 310

4. In how many different ways can the letters of the word “SALTY” be arranged?

Correct Answer: (e) 120
Solution:

The number of ways = 5! = 120

5. There are 7 periods in each working day of a college. In how many ways can one organise 6 subjects such that each subject is allowed at least one period?

Correct Answer: (b) 15120
Solution:

6 subjects can be arranged in periods in ⁷P₆ ways.
Remaining 1 period can be arranged in ⁶P₁ ways.
Two subjects alike in each of the arrangement.
So we need to divide by 2! to avoid over counting.

Total number of arrangements
= (⁷P₆ × ⁶P₁)/2!
= 5040 × 6 / 2
= 30240 / 2
= 15120

6. Among a set of 5 blue balls and 7 red balls, how many selections of 5 balls can be made such that at least 3 of them are blue balls?

Correct Answer: (e) 246
Solution:

Combinations: 3 blue and 2 red, 4 blue and 1 red, 5 blue and 0 red
Hence, ⁵C₃ × ⁷C₂ + ⁵C₄ × ⁷C₁ + ⁵C₅ × ⁷C₀ = 10 × 21 + 5 × 7 + 1 = 246 ways

7. How many ways the letters of the word “CURRENT” be arranged?

Correct Answer: (a) 2520  
Solution:Required ways = 7!/2! = 2520

8. In how many ways a selection of 4 students having at least 2 boys can be selected from 4 boys and 5 girl students?

Correct Answer: (d) 81  
Solution:

Number of ways = ⁵C₂ × ⁴C₂ + ⁴C₃ × ⁵C₁ + ⁴C₄
= 10 × 6 + 4 × 5 + 1
= 60 + 20 + 1 = 81

9. In how many ways can a committee of 4 people be chosen out of 8 people?

Correct Answer: (c) 70
Solution:Required number of ways = 8 C 4 = 70

10. In how many ways the letters of word “SIMPLE” can be arranged so that all vowels come together?

Correct Answer: (c) 240
Solution:

Total letters = 6
Vowels = 1,
E = 2
Required ways = 2! × 5! = 240