## Rajasthan Board RBSE Class 11 Maths Chapter 6 Permutations and Combinations Miscellaneous Exercise

Question 1.

If ^{n}P_{n-2} = 60, then n will be:

(A) 2

(B) 4

(C) 5

(D) 3

Solution:

Question 2.

^{n}P_{r} ÷ ^{n}C_{r} is equal to:

(A) n!

(B) (n – r)!

(C) \(\frac { 1 }{ r! }\)

(D) r!

Solution:

Hence, the option (D) is correct.

Question 3.

How many ways 5 persons can sit around the round table:

(A) 120

(B) 24

(C) 60

(D) 12

Solution:

Total number of ways to sitting 5 persons around the round table

= (5 – 1)! = 4!

= 4 × 3 × 2 × 1 = 24.

Hence, option (B) is correct.

Question 4.

How many words can be formed using letters BHILWARA:

(A) \(\frac { 8! }{ 2! }\)

(B) 8!

(C) 7!

(D) \(\frac { 6! }{ 2! }\)

Solution:

There are 8 letters in the given word in which 2 A’s and other letters are different.

Then numbers of words formed by letters of BHILWARA = \(\frac { 8! }{ 2! }\)

Hence option (A) is correct.

Question 5.

Solution:

Question 6.

Find the value of ^{61}C_{57} – ^{60}C_{56}:

(A) ^{61}C_{58}

(B) ^{60}C_{57}

(C) ^{60}C_{58}

(D) ^{60}C_{56}

Solution:

Question 7.

If ^{15}C_{3r} = ^{15}C_{r+3}, then r is equal to:

(A) 5

(B) 4

(C) 3

(D) 2

Solution

Question 8.

There are 6 points on the circumference of a circle, the number of straight lines joining their points will be:

(A) 30

(B) 15

(C) 12

(D) 20

Solution:

Number of lines passing through n points = ^{n}C_{2}

Number of lines passing through 6 points

Hence, option (B) is correct.

Question 9.

How many words can be formed using letters of BHOPAL?

(A) 124

(B) 240

(C) 360

(D) 720

Solution:

Here, the number of letters are 6 and every time we take 6 letters.

Hence, required number of = ⌊6 = 6 × 5 × 4 × 3 × 2 × 1 = 720

Hence, option (D) is correct.

Question 10.

There are 4 points on the circumference of a circle by joining them how many triangles can be formed?

(A) 4

(B) 6

(C) 8

(D) 12

Solution:

There are three vertices in a triangle.

Given: There are 4 points on the circumference of a circle.

Then, the number of required triangles

Hence, option (A) is correct.

Question 11.

If ^{n}C_{2} = ^{n}C_{7}, then find ^{n}C_{16}

Solution:

Question 12.

Find the value of n:

(i) ^{n}C_{2} : ^{n}C_{2} = 12 : 1

(ii) ^{2n}C_{3} : ^{n}C_{3} = 11 : 1

Solution:

Question 13.

Find the number of chords passing through 11 points on the circumference of a circle.

Solution:

Number of the point on the circumference of circle = 11

We know that a chord is formed by joining two points.

Hence taking 2 points out of 11 points number of chords = ^{11}C_{2}

Hence, the number of chords = 55.

Question 14.

Determine the number of combinations out of deck of 52 cards of each selection of 5 cards has exactly one ace.

Solution:

Number of cards in a deck = 52

Total number of aces = 4

Hence Number of remaining cards = 52 – 4 = 48

We have to make a collection of 5 cards, in which there is 1 ace and 4 other cards.

Hence, number of ways choosing 1 out of 4 ace.

Now,Number of ways choosing 4 cards out of 48 cards

Total number of collection having 5 cards

= ^{4}C_{1} × ^{48}C_{4} = 4 × 194580 = 778320

Hence, number of combinations formed by 5 cards = 778320.

Question 15.

There are n points in a plane in which m points are collinear. How many triangles will be formed by joining three points?

Solution:

For making a triangle we need three points thus if 3 points out of n points are not in a line, then ^{n}C_{3} triangle can be formed by n points but m point is in a line, thus ^{m}C_{3} triangle forms less.

Hence, required number of triangles = ^{n}C_{3} – ^{m}C_{3}.

Question 16.

Find the number of diagonals of a decagon.

Solution:

There are 10 vertices in a decagon. Now by joining two vertices, we get a side of decagon or a diagonal.

Hence, the number of the line segment by joining of vertices of decagon = number of ways taking 2 points out of 10 points

Hence, the number of diagonals is 35.

Question 17.

There are 5 empty seats in a train, then in how many ways three travellers can sit on these seats?

Solution:

A number of ways of seating of 3 travellers out of 5 seats.

Hence, 3 travellers can be seated in 60 ways.

Question 18.

A group of 7 has to be formed from 6 boys and 4 girls. In how many ways a group can be formed if boys are in the majority in this group?

Solution:

Number of boys = 6

Number of girls = 4

Number of members in the group = 7

One putting boys in the majority in the group

We have to choose in the following ways :

4 boys + 3 girls

5 boys + 2 girls

6 boys + 1 girls

Number of ways choosing 4 boys and 3 girls = ^{6}C_{4} × ^{4}C_{3}

Number of ways choosing 5 boys and 2 girls = ^{6}C_{5} × ^{4}C_{2}

Number of ways choosing 6 boys and 1 girl = ^{6}C_{6} × ^{4}C_{1}

Required number of ways forming the group

Hence, the group can be formed in 100 ways.

Question 19.

In the conference of 8 persons, every person handshake with each other only once, then find the total number of hand shook.

Solution:

When two persons shake hand with each other, then this is taken as one handshake.

Hence, the number of handshakes is equal to choosing 2 persons out of 8 persons.

Number of handshake

Hence, the required number of hand shaken is 28.

Question 20.

In how many ways 6 men and 6 women can sit around the round table whereas any two women never sit together?

Solution:

First, we sit the men in an order that there is a place vacant between two men.

Then, number of ways sitting 6 men = (6 – 1)! = 5!

Now a number of ways sitting 6 women in the vacant places = 6!

Number of permutations = (6 – 1)! = 5!

Now, the number of ways sitting 6 women in the vacant place = 6!

Number of permutaions = 5! × 6!

= (5 × 4 × 3 × 2 × 1) × (6 × 5 × 4 × 3 × 2 × 1)

= 86400.

Question 21.

In how many ways can the letters of the word ASSASSINATION be arranged so that all the S are together?

Solution:

Number of letters in word ASSASSINATION = 13

In this word A comes 3 times, S comes 4 times, I comes 2 times, N comes 2 times and other letters are different.

Now, taking 4’s together, then consider it as one letter, now adding other 9 letters and thus the permutation formed by 10 letters

= \(\frac { 10! }{ 2!2!3! }\)

[Because I comes 2 times, N comes 2 times and A comes 3 times]

Hence, the total number of ways is 151200.