BANK & INSURANCE (PERMUTATION AND COMBINATION) PART 1

Total Questions: 30

21. In a language, there are six different words. A sentence can be formed by at least 2 words. If order of words is changed in a sentence, we get a different sentence. How many different sentences can be formed in this language?

Correct Answer: (c) 1950
Solution:

Here, different order gives different sentence.
So, permutations are needed to make sentences.
⇒ Different sentences that can be formed
= ⁶P₂ + ⁶P₃ + ⁶P₄ + ⁶P₅ + ⁶P₆
= 30 + 120 + 360 + 720 + 720 = 1950

22. How many 3-letter words with or without meaning, can be formed out of the letters of the word, ‘LOGARITHMS’, if repetition of letters is not allowed?

Correct Answer: (a) 720  
Solution:The word “LOGARITHMS” has 10 different alphabets
Hence, the number of 3-letter words (with or without meaning) formed by using these letters
= (10)P₃ = 10 × 9 × 8 = 720

23. There are 2 shirts, 3 jeans, 3 socks and 2 skirts. In how many ways a shopkeeper can arrange these things so that all the socks come together and all the skirts come together?

Correct Answer: (b) 5040  
Solution:

We will count 3 socks as 1 socks and 2 skirts as 1 skirt
Total ways = 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1
= 5040

24. In how many different ways can the letters of the word WINDOW be arranged in such a way that the vowels never come together?

Correct Answer: (d) 300  
Solution:

Total ways = 360 ways
when all the vowels always come together = 60
Ways when all the vowels never come together
= 360 − 60 = 300

25. The number of ways in which 8 different books can be arranged on a shelf so that 3 particular books shall not be together:

Correct Answer: (d) 36000  
Solution:Number of ways in which 8 books can be arranged = 8!
Number of ways when three particular books are together = 6! × 3!
Therefore Number of ways when three particular books are not together
= 8! − 6! × 3!
= 6!(7 × 8 − 3 × 2) = 6! × 50
= 720 × 50 = 36000

26. In a room everybody shakes hands with everybody else. The total number of hand-shakes is 66. The total number of persons in the room is:

Correct Answer: (d) 12  
Solution:

Let the total number of persons be N.
Given, total number of hand-shakes is 66
For a hand shake we require two people, total number of handshake is ⁿC₂
∴ ⁿC₂ = 66
∴ N(N−1)/2 = 66
⇒ N² − N = 132
⇒ N² − N − 132 = 0
⇒ (N−12)(N+11) = 0
⇒ N = 12 persons

27. How many numbers are there in between 100 and 1000 such that exactly one of their digits is 3 if repetition is not allowed?

Correct Answer: (b) 200  
Solution:

Surely 3 can occur at either hundreds place or tens place or units place. So three cases arise.

(a) If 3 occurs at hundreds place then the digit at tens place can be chosen in only nine ways (all ten digits leaving only 3 so we are left with 9 digits) and digit at units place can be chosen in only 8 ways (as 3 and digit at tens place cannot be used again)
So total such numbers = 1 × 9 × 8 = 72

(b) If 3 occurs at tens place then its hundreds place can be chosen in only 8 ways (because use of 3 is not allowed and if we use 0 out of the remaining 9 digits it will be a 2-digit number which is not allowed) and unit place can be chosen only in 8 ways (since digit at hundreds place and 3 is not allowed)
So total such numbers = 8 × 1 × 8 = 64

(c) If 3 occurs at units place then its hundreds place can be chosen in only 8 ways (because use of 3 is not allowed and if we use 0 out of the remaining 9 digits it will be a 2-digit number which is not allowed) and tens place can be chosen only in 8 ways (since digit at hundreds place and 3 is not allowed)
So total such numbers = 8 × 8 × 1 = 64

Hence total such numbers = 72 + 64 + 64 = 200

28. In how many ways a word NEWSPAPER can be arranged so that N, S and W always come together?

Correct Answer: (d) 1260  
Solution:

Total Words = N, S, W, E, P, A, P, E, R = 7
Total ways = 7! ÷ (2! × 2!)
= 7 × 6 × 5 × 4 × 3 × 2 × 1 ÷ 2 ÷ 2
= 1260

29. In how many ways a group of 3 students can be selected from 7 men and 5 women consisting of 1 man and 2 women?

Correct Answer: (d) 70  
Solution:No. of ways of selecting one man out of 7 women
= ⁷C₁ = 7
No of ways of selecting 2 women out of 5 women
= ⁵C₂ = 10
Required ways = ⁷C₁ × ⁵C₂ = 7 × 10 = 70

30. In an auditorium the chairs were arranged such that the number of rows were 3 more than the number of columns. The chairs are rearranged by removing 3 columns and adding 6 rows without adding or removing any chair. How many people can sit in that auditorium at a time?

Correct Answer: (c) 108
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 − 3)(x + 9)
∴ x² + 3x = x² + 6x − 27
∴ x = 9
Thus there were 9 columns and 12 rows, i.e., 9 × 12
= 108 people can sit in the auditorium at a time.