BANK & INSURANCE (PERMUTATION AND COMBINATION) PART 2

Total Questions: 30

1. In how many ways can five boys be made to stand in a row such that two of them, P and Q are always together?

Correct Answer: (b) 48  
Solution:

If P and Q are always together, then the total number of boys is 4.
No. of ways = 4!
P and Q also interchange their place.
No. of ways = 2!
Total ways = 4! × 2! = 48

2. In how many ways the letters of the word “WORDART” be arranged?

Correct Answer: (d) 2520  
Solution:

Required number of ways = 7!/2! = 2520

3. In how many number of ways the word “TYPICAL” be arranged?

Correct Answer: (a) 5040  
Solution:

Required number of ways = 7! = 5040

4. In an auditorium the chairs were arranged such that the number of rows was 3 more than the number of columns. The chairs are rearranged by removing 4 columns and adding 8 rows without adding or removing any chair. How many people can sit in that auditorium at a time?

Correct Answer: (b) 154  
Solution:Since no chair was added or removed, the capacity of the auditorium remains constant.
The capacity of the auditorium is the product of the number of rows and number of columns.
Let there be x columns and x + 3 rows, then
⇒ x (x + 3) = (x − 4)(x + 3 + 8)
⇒ x (x + 3) = (x − 4)(x + 11)
⇒ x² + 3x = x² + 11x − 4x − 44
⇒ 4x = 44
⇒ x = 11
Therefore, there were 11 columns and 14 rows,
So, 11 × 14 = 154 people can sit in the auditorium at a time

5. In how many ways word “ENERGY” be arranged in that all vowels and consonants come together?

Correct Answer: (a) 48  
Solution:Number of vowels = E, E = 2
Number of consonants = N, R, G, Y = 4
Number of ways = (2! × 4! × 2!)/2! = 48

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

Correct Answer: (e) None of these
Solution:Number of ways = ⁵C₂ × ⁴C₂ + ⁵C₁ × ⁴C₃ + ⁴C₄
= 10 × 6 + 5 × 4 + 1
= 60 + 20 + 1 = 81

7. In how many ways the word “PRIDE” be arranged so that all vowels and consonants come together?

Correct Answer: (c) 24
Solution:

Number of vowels = 2
Number of consonants = 3
Number of ways = 2! × 3! × 2! = 24

8. In how many ways a committee of 5 members can be formed from 6 men and 7 women in which at least 3 men should come?

Correct Answer: (a) 531  
Solution:

Possible selection = 3 men + 2 women, 4 men + 1 woman, 5 men
= ⁶C₃ × ⁷C₂ + ⁷C₄ × ⁷C₁ + ⁶C₅
= (6×5×4/1×2×3)(7×6/1×2) + (6×5/1×2) ×7+6
= 20×21 + 15×7 + 6
= 531

9. How many 3 digit numbers can be formed from the digit 1, 2, 3, 5, 7 which are divisible by 5 and none of the digits is repeated?

Correct Answer: (d) 12  
Solution:

Number is divisible by 5.
Hence third digit should be 5.
Remaining 1, 2, 3, 7 = 4 digits
Second digit can be filled in 4 number of ways and
First digit can be filled in 3 number of ways.
Total required ways = 3 × 4 × 1 = 12

10. How many words of 4 consonants and 4 vowels can be formed, out of 8 consonants and 5 vowels?

Correct Answer: (b) 350 × 8!
Solution:No. of ways to choose 4 consonants out of 8 consonants = ⁸C₄
No. of ways to choose 4 vowels out of 5 vowels = ⁵C₄
These 8 letters can be arranged in 8! ways.
Required number of words = ⁸C₄ × ⁵C₄ × 8! = 350 × 8!