BANK & INSURANCE (PERMUTATION AND COMBINATION) PART 3

Total Questions: 30

1. In how many ways can the letters of the word “PARAGLIDING” be arranged such that all the vowels occur together?

Correct Answer: (b) 120960 ways
Solution:In the word “PARAGLIDING” there are 11 letters in which there are 4 vowels (i.e. 2 A’s and 2 I’s) and 7 consonants (i.e 2 G’s and each of P, R, L, D, N)
Considering vowel as one letter, the number of letters becomes 8 which can be arranged as
8! / 2! = 40320 / 2 = 20160
Vowel A and I appear twice, so vowels can be arranged as
4! / (2! × 2!) = 24 / 4 = 6
Hence the required number of ways in which the letters of the word “PARAGLIDING” can be arranged so that all the vowels occur together
= 20160 × 6 = 120960
Hence, option b is correct.

2. Five people out of whom only two can drive are to be seated in a five seater car with two seats in front and three in the rear. The people who know driving don’t sit together. Only someone who knows driving can sit on the driver’s seat. Find the number of ways the five people can be seated.

Correct Answer: (d) 36  
Solution:Number of people who can drive = 2
Number of ways of selecting driver = ²C₁
The other person who knows driving can be seated only in the rear three seats in 3 ways
Total number of ways of seating the two persons = ²C₁ × 3
Number of ways of seating remaining = 3!
Total number of all five can be seated = ²C₁ × 3 × 3! = 36
Hence, correct answer is 36
Hence, option d is correct.

3. A boy is playing a Snake & ladder game; he is on 91 and has to get to 100 to complete the game. There is a snake on 93 and 96. In how many ways he can complete the game, if he doesn’t want to roll the dice more than three times.

Correct Answer: (c) 16
Solution:

91 — 92 — 93 — 94 — 95 — 96 — 97 — 98 — 99 — 100
Total position advance needed = 100 − 91 = 9
One roll of dice can’t complete the game.

If he completes in two roll of dice.
Possible dice throws are – (3&6), (4&5), (5&4), (6&3)
But (5&4) will bring the token on 96, so this is rejected.

If he completes the game in three roll of dices
First dice reading options are 1,3,4,6
After checking all option and rejecting those in which token reaches on 93 or 96
Possible dice throws are (1,2,6), (1,3,5), (1,5,3), (1,6,2); (3,1,5), (3,3,3), (3,4,2), (3,5,1); (4,2,3), (4,3,2), (4,4,1), (6,1,2), (6,2,1)
Total number of ways = 16
Hence, option c is correct

4. 8 members are to be selected from a group of 9 males and 7 females. In how many ways will the members with at most 3 females and at least 4 males be selected?

Correct Answer: (c) 6435 ways  
Solution:

Case I: 5 males and 3 females can be selected
Number of ways of selection = ⁹C₅ × ⁷C₃
= 126 × 35 = 4410

Case II: 6 males and 2 females can be selected
Number of ways of selection = ⁹C₆ × ⁷C₂
= 84 × 21 = 1764

Case III: 7 males and 1 female can be selected
Number of ways of selection = ⁹C₇ × ⁷C₁
= 36 × 7 = 252

Case IV: 8 males can be selected Number of ways of selection = ⁹C₈ = 9

So, total number of ways of selecting the members = 4410 + 1764 + 252 + 9 = 6435 ways
Hence, option c is correct.

5. A chess board has rows and columns marked A to H and 1-8. Aman has a knight and a rook which he has to place on the board such that the two pieces are not in same row or column, what is total number of ways he can place the two pieces?

Correct Answer: (b) 3136  
Solution:

Let us select any box out of 64 for placing knight,
no of ways = ⁶⁴C₁

Now, row 6 and column c can’t be used to place rook.
Remaining boxes = 64 − (8 + 7) = 49
The rook can be place in any of 49 boxes,
no of ways = ⁴⁹C₁
Total number of possible ways = ⁴⁹C₁ × ⁶⁴C₁
= 3136
Hence, option b is correct.

6. How many three letter words can be formed using the letters of the word “PRACTICES”?

Correct Answer: (d) 357  
Solution:

Combinations of three different letter

Number of combinations

Number of permutations for each combination

Total number of permutations

3 different letter

8C3 = 56

6

336

2 same letter (eg. c.c.v)

7C1 = 7

3

21

Total = 357

Hence, option D is correct.

7. Six students sitting in a row are given one toffee each from three types of toffees such that no two adjacent child gets same type of toffee. In how many ways can the toffees be distributed among the students?

Correct Answer: (c) 96
Solution:Let the students be S1,S2,S3,S4,S5,S6 and A, B and C be three types of coffee
S1 can get any of the 3 from A, B, and C in 3 ways
S2 can get any of the 2, other than what A got in 2 ways
S3,S4,S5,S6 each can get different toffee in 2 ways
Total numbers of ways in which distribution can be done 3 × 2 × 2 × 2 × 2 × 2 = 96
Hence, option c is correct.

8. In how many different ways can the letters of the word “Thoughts” be arranged in such a way that the vowels always come together?

Correct Answer: (c) 2520 ways  
Solution:In the word “Thoughts”, there are 2 vowels O and U and 6 consonants, 2T’s, 2H, 1G and 1S.
Number of ways = 7! × 2! / (2! × 2!)
⇒ 35 × 9 × 8 = 2520 ways.
Hence, option C is correct

9. An objective test with all the questions mandatory to be answered can be attempted in 127 ways such that the student gets at least one question right. Find the number of ways in which he can answer 4 questions correctly

Correct Answer: (b) 35
Solution:Any question can be answered in 2 ways (right or wrong)
Let the number of questions be N
2ᴺ − 1 = 127
Therefore N = 7
Number of ways in answering 4 answers correctly
= ⁷C₄ = 35
Hence, option b is correct

10. A postmaster wants to get delivered 6 letters at six different addresses. In the post office there are 2 postmen then in how many ways can the postmaster send the letters at different addresses through the postmen?

Correct Answer: (c) 64 
Solution:Each letter can be delivered at the six different addresses in 2 different ways
Hence, the required number of ways = 2⁶ = 64
Hence, option c is correct.