BANK & INSURANCE (PERMUTATION AND COMBINATION) PART 3

Total Questions: 30

21. Directions (Set of 2 Questions): Kindly study the following information carefully and answer the question that follows:

Five students are to be arranged on five chairs for a photograph. Three of these are girls and the rest are boys.

Ques : Find out the total number of ways in which three girls are together.

Correct Answer: (a) 36  
Solution:

As per the question, three girls can’t occupy consecutive seats but two can. Therefore, if we find the number of ways in which all three girls occupy consecutive seats and subtract this number from the total number of ways in which the five people can be arranged among themselves, we will get the required answer.

5 students can be arranged among themselves in ⁵P₅ ways
= 120 ways.

Assume that the 3 girls are one entity.
The total number of ways in which they can be arranged among themselves
= 3! = 6

Also, the set of three girls and the other students can be arranged among themselves in 3! = 6 ways.

Thus, total number of ways in which three girls are together
= 6 × 6 = 36

Thus, number of ways in which all 3 girls will not occupy consecutive seats = 120 − 36 = 84

As per the common explanation, we get
Total number of ways in which three girls are together = 6 × 6 = 36
Hence, option a is correct.

22. Find out the number of ways in which all three girls do not occupy consecutive seats.

Correct Answer: (c) 84
Solution:Following the common explanation, we get
Thus, number of ways in which all 3 girls will not occupy consecutive seats = 120 − 36 = 84
Hence, option c is correct

23. Directions (Set of 2 Questions): Kindly study the following information carefully and answer the question that follows:

Using all the letters of the word LINEAR.

Ques: How many words start with a vowel but end with a consonant?

Correct Answer: (d) 216  
Solution:

The word LINEAR has three vowels - I, E and A.
If a word starts and ends with a vowel, the two letters to occupy the first and the last positions can be selected and arranged in ³P₂ = 6 ways.
The remaining 4 letters can be arranged among themselves in ⁴P₄ = 4! = 24 ways.
∴ The number of words that start and end with a vowel = 24 × 6 = 144.

If a word starts with a vowel but ends with a consonant, its first letter can be selected from I, E and A in 3 ways.
Its last letter can be selected from L, N and R in 3 ways.
The remaining three letters can be arranged in 4! ways.
∴ The number of words that start with a vowel but end with a consonant = 3 × 3 × 4! = 9 × 24 = 216.

Following the common explanation, we get
The number of words that start with a vowel but end with a consonant = 9 × 24 = 216.
Hence, option D is correct.

24. How many different words can be formed that start and end with vowel?

Correct Answer: (c) 144
Solution:Following the common explanation, we get
The number of words that start and end with a vowel = 144.
Hence, option c is correct.

25. A six letter word is to be formed by using at least two vowels in it. How many such words can be formed (not necessarily meaningful) if all the letters in word are different?

Correct Answer: (b) 53439120  
Solution:

Six letter words with at least two vowels can have 2, 3, 4 or 5 vowels as no letters can be repeated.
There are 21 consonants and 5 vowels.
All possible cases: 2 vowels and 4 consonants 3 vowels and 3 consonants 4 vowels and 2 consonants 5 vowels and 1 consonant
∴ Number of ways in which this can be done
= ⁵C₂ × ²¹C₄ + ⁵C₃ × ²¹C₃ + ⁵C₄ × ²¹C₂ + ⁵C₅ × ²¹C₁
= 10 × 5985 + 10 × 1330 + 5 × 210 + 1 × 21
= 74221

In each of these cases, chosen 6 letters can arrange themselves in 6! ways.
∴ Total number of ways in which this can be done
= 6! × 74221
= 720 × 74221 = 53439120
Hence, option b is correct.

26. In a badminton competition involving some men and women of a society, every person had to play exactly one game with every other person. It was found that in 36 games both the players were men and in 78 games both the players were women. Find the number of games in which one player was a man and another was a woman.

Correct Answer: (b) 117  
Solution:

Let the number of men be x and women be y
In badminton two person can play at a time,
Therefore, no of games played between men is
= ˣC₂ = 36

x(x − 1) / 2 = 36
x(x − 1) = 72
x = 9

Which means total number of men playing badminton are 9

Now, no of games played between women is
= ʸC₂ = 78

y(y − 1) / 2 = 78
y(y − 1) = 156
y = 13

Which means total number of women playing badminton are 13

Therefore, no of games in which one player is man and one is woman is,
= ⁹C₁ × ¹³C₁ = 117
Hence, option (b) is correct.

27. In how many ways can 5 prizes be distributed to 8 students if each student can get any number of prizes?

Correct Answer: (c) 8⁵
Solution:Number of ways = 8⁵

28. In how many different ways letters of the word “EDUCATION” can be arranged such that all the consonants come together?

Correct Answer: (c) 17280
Solution:Number of consonants = 4
Consonants can be arranged among themselves in 4! Ways.
E, U, A, I, O and (DCTN) can be arranged in 6! Ways
Required number of ways = 6! × 4! = 720 × 24 = 17280
Hence, option (C) is correct.

29. Directions (Set of 2 Questions): Kindly study the following information carefully and answer the question that follows:

There are 5 singers, 8 dancers and 7 actors in a movie. A group of 6 members will be chosen to nominate in an award show.

Ques: How many combinations of members are possible if the group is to consist of all members of the same profession?

Correct Answer: (d) 35  
Solution:

Here we want 6 singers or 6 dancers or 6 actors.
The group cannot be of all singers since there are only 5 singers.
Therefore, the group can be a group of 6 dancers or 6 actors.

Number of groups of dancers = ⁸C₆ = 28
Number of groups of actors = ⁷C₆ = 7

Since we want the number of groups of 6 dancers or 6 actors,
we want the sum of each of these possibilities:
= 28 + 7 = 35
Hence, option (d) is correct.

30. How many combinations of members are possible if the group is to consist of exactly 3 singers?

Correct Answer: (c) 4550
Solution:

Here we need the number of possible combinations of 3 out of 5 singers, ⁵C₃, and the number of possible combinations of 3 out of the 15 dancers and actors, ¹⁵C₃.
Note that we want 3 singers and 3 members from the other profession. Therefore, we multiply the number of possible groups of 3 of the 5 singers times the number of possible groups of 3 of the 15 members from the other profession.

⁵C₃ × ¹⁵C₃ = 4550
Hence, option (c) is correct