Gujarat Board Statistics Class 11 GSEB Solutions Chapter 6 Permutations, Combinations and Binomial Expansion Ex 6.2 Textbook Exercise Questions and Answers.

## Gujarat Board Textbook Solutions Class 11 Statistics Chapter 6 Permutations, Combinations and Binomial Expansion Ex 6.2

Question 1.

Obtain the values of the following:

(1) ^{11}C_{4}

Answer:

^{n}C_{r} = \(\frac{n !}{r !(n-r) !}\)

∴ ^{11}C_{4} = \(\frac{11 !}{41(11-4) !}\)

= \(\frac{11 \times 10 \times 9 \times 8 \times 7 !}{4 \times 3 \times 2 \times 1 \times 7 !}\)

= \(\frac{7920}{24}\) = 330

(2) ^{9}C_{0}

Answer:

^{n}C_{0} = 1

∴ ^{9}C_{0} = 1

(3) ^{25}C_{23}

Answer:

^{n}C_{r} = \(\frac{n !}{r !(n-r) !}\)

∴ ^{25}C_{23} = \(\frac{25 !}{23 !(25-23) !}\)

= \(\frac{25 !}{23 ! 2 !}\)

= \(\frac{25 \times 24 \times 23 !}{23 ! \times 2 \times 1}=\frac{600}{2}\) = 300

(4) ^{8}C_{8}

Answer:

^{n}C_{r} = 1

∴ ^{8}C_{8} = 1

Question 2.

Find the unknown value:

(1) ^{n}C_{2} = 28

Answer:

^{n}C_{2} = 28

∴ \(\frac{n !}{2 !(n-2) !}\) = 28

∴ \(\frac{n(n-1)(n-2) !}{2 \times(n-2) !}\) = 28

∴ n(n – 1) = 28 × 2

∴ n(n – 1) = 56

∴ n(n – 1) = 8 × 7

∴ n(n – 1) = 8(8 – 1)

∴ n = 8

(2) ^{27}C_{r+4} = ^{27}C_{2r-1}

Answer:

^{27}C_{r+4} = ^{27}C_{2r-1}

^{n}C_{x} = ^{n}C_{y}

Or

n = x + y

^{27}C_{r+4} = ^{27}C_{2r-1}

∴ r + 4 = 2r – 1

∴ 4 + 1 = 2r – r

∴ r = 5

Or

r + 4 + 2r – 1 = 27

∴ 3r + 3 = 27

∴ 3r = 27 – 3

∴ r = \(\frac{27}{3}\) = 8

(3) ^{n}C_{n-2} = 15

Answer:

∴ \(\frac{n !}{(n-2) !(n-n+2) !}\) = 15

∴ \(\frac{n(n-1)(n-2) !}{(n-2) ! 2 !}\) = 15

∴ \(\frac{n(n-1)}{2}\) = 15

∴ n(n – 1) = 30

∴ n(n – 1) = 6 × 5

∴ n(n – 1) = 6(6 – 1)

∴ n = 6

(4) 4.^{n}C_{4} = 7.^{n}C_{3}

Answer:

Question 3.

8 candidates applied for 2 posts of peon in a school. In how many ways can 2 peons be selected from 8 candidates?

Answer:

Total numbcr of combinations selecting 2 candidates for the post of peon out of 8 candidates

= ^{8}C_{2}

= \(\frac{8 !}{2 !(8-2) !}\)

= \(\frac{8 !}{2 ! 6 !}\)

= \(\frac{8 \times 7 \times 6 !}{2 \times 6 !}\)

=\(\frac{8 \times 7}{2}\) = 28

Alternative method:

^{8}C_{2} = \(\frac{8 \times 7}{2 \times 1}\)

= 28

Question 4.

5 countries participate In a cricket tournament. In the first round, every country plays a match with the other country. How many matches will be played in this round?

Answer:

5 countries participate In a cricket tournament. Every country plays a match with the other country. Therefore, In every round two countries are selected to play a match.

∴ Total combinations = ^{5}C_{2}

= \(\frac{5 \times 4}{2 \times 1}\)

= 10 Matches

Question 5.

There are 200 items in a box and 5 % of them are defective. In how many ways can 3 items can be selected from the box so that all the items selected are defective?

Answer:

5 % of items are defective in a box of 200 items.

∴ No. of defective items in the box = 200 × \(\frac{5}{100}\) =10 items.

3 items are selected from the box. Such that all three items are defective.

∴ Total combinations = ^{10}C_{2}

= \(\frac{10 \times 9 \times 8}{3 \times 2 \times 1}\)

= \(\frac{720}{6}\) = 120

Question 6.

In how many ways can 3 clerks and 1 peon be selected from 14 clerks and 6 peons working in a bank?

Answer:

3 clerks out of 14 clerks can be selected in ^{14}C_{3} ways and one peon out of 6 peons can be selected in ^{6}C_{1} ways.

∴ Total combinations = ^{14}C_{3} × ^{6}C_{1}

= \(\frac{14 \times 13 \times 12}{3 \times 2 \times 1}\) × 6

= 2184

Question 7.

There are 3 white and 5 pInk flowers in a box. In how many ways can

(1 ) three flowers of same colour be selected?

(2) 2 flowers of different colours be selected?

Answer:

In a box there are 3 white and 5 pInk flowers.

(1) Selected three flowers are of the same colour:

3 whIte flowers of 3 white flowers can be selected in ^{3}C_{3} ways or

3 pink flowers of 6 pInk flowers can be selected in ^{6}C_{3} ways.

∴ Total combinations = ^{3}C_{3} + ^{5}C_{3}

= 1 + \(\frac{5 \times 4 \times 3}{3 \times 2 \times 1}\)

= 1 + 10 = 11

(2) Selected two flowers are of different colours:

1 white flowers out of 3 whIte flowers can be selected in 3C1 ways and

1 pink flower out of 5 pInk flowers can be selected In 5C1 ways.

∴ Total combinations = ^{1}C_{1} × ^{8}C_{3}

= 3 × 5

= 15

Question 8.

Two cards are randomly selected from a pack of 52 cards. In how many ways can 2 cards be selected such that,

(1) both are of heart?

(2)one Is a king and the other is a queen?

Answer:

(1) In a pack of 52 cards, there are 13 cards of heart. 2 cards of heart out of 13 cards of heart can be selected in ways.

∴ Total combinations = ^{13}C_{2} × ^{6}C_{3}

= \(\frac{13 \times 12}{2}\)

= 20

(2) In a pack of 52 cards, there are 4 cards of king and 4 cards of queen one card of king out of 4 cards of king can be selected in ^{4}C_{1} ways and one card of queen out of 4 cards of queen can be selected In ^{4}C_{1} ways.

∴ Total combinations = ^{4}C_{1} × ^{4}C_{1}

= 4 × 4

= 16

Question 9.

There are 9 employees in a bank of whIch 6 are clerks, 2 are peons and 1 is a manager. In how many ways can a committee of 4 members be formed such that

(1 ) the manager must be selected?

(2 ) two peons are not to be selected and the manager Is to be selected?

Answer:

Out of 9 employces In a bank. 6 are clerks. 2 are peons and 1 Is a manager. A committee of 4 members Is to be formed.

(1) The manager must be selected In the committee:

If the manager Is to be selected In the committee of 4 members, than from the remaining (6 clerks + 2 peons) 8 employees, the remaining 3 members of the committee can be selected in ^{8}C_{3} ways.

∴ Total combinations = ^{1}C_{1} × ^{8}C_{3}

= 1 × \(\frac{8 \times 7 \times 6}{3 \times 2 \times 1}\)

= 56

(2) Two peons are not be selected and the manager is to be selected:

If the manager Is to be selected In the committee of 4 mcmbcrs and two peons are not be selected than from the remaIning 6 clerks, the remaining 3 members of the committee can be selected in^{6}C_{3} ways.

∴ Total combinations = ^{1}C_{1} × ^{6}C_{3}

= 1 × \(\frac{6 \times 5 \times 4}{3 \times 2 \times 1}\)

=20

Question 10.

In an office, there are 8 employees of which 3 are females and remaining are males. 3 employees are to be selected from the office for training. In how many ways can the selection be done so that at least one male Is selected?

Answer:

In an office, out of 8 employees 3 are females and 5 are males. 3 employees are to be selected for training.

The different options for selecting 3 employees so that at least one male is selected are as follows:

- One male and 2 females OR
- Two males and 1 female OR
- Three males and no female

Question 11.

A person has 6 friends. In how many ways can he invite at least one friend to his house?

Answer:

A person has 6 friends. He wants to invite at least one of his friends to his house.

He may invite his 1 friend or 2 friends or 3 friends or 4 friends or 5 friends or 6 friends out of his 6 friends.

∴ Total combinations

= ^{6}C_{1} + ^{6}C_{2} + ^{6}C_{3} + ^{6}C_{4} + ^{6}C_{5} + ^{6}C_{6}

= 6 + \(\frac{6 \times 5}{2 \times 1}+\frac{6 \times 5 \times 4}{3 \times 2 \times 1}+\frac{6 \times 5 \times 4 \times 3}{4 \times 3 \times 2 \times 1}\) + 6 + 1

= 6 + 15 + 20 + 15 + 6 + 1 = 63

Question 12.

In how many ways can 5 books be selected from 8 different books so that.

(1) a particular book is always selected?

(2) a particular books is never selected?

Answer:

Out of 8 different books, 5 books are to be selected.

(1) A particular book is always selected : If a particular book is always selected, then out of 7 remaining books, the remaining 4 books can be selected in ^{7}C_{4} ways.

∴ Total combinations = ^{1}C_{1} × ^{7}C_{4}

= 1 × \(\frac{7 \times 6 \times 5 \times 4}{4 \times 3 \times 2 \times 1}\)

= 35

(2) A particular book is never selected:

If a particular book is never selected, then out of the remaining 7 books, 5 books can be selected in 7C5 ways.

∴ Total combinations = ^{7}C_{5}

= \(\frac{7 \times 6 \times 5 \times 4 \times 3}{5 \times 4 \times 3 \times 2 \times 1}\)

= 21

Question 13.

A student in 12th standard commerce stream has to appear for exam in 7 subjects. It is necessary to pass in all the subjects to pass an exam. Certain minimum marks must be obtained to pass in a subject. In how many ways can a student appearing for the exam fail?

Answer:

A student has to appear for exam in 7 subjects. Certain minimum marks are required to pass in the subject. He fails if he does not reach to the level of minimum marks required in at least one of 7 subjects.

Total combinations in which a student can fail

= ^{7}C_{1} + ^{7}C_{2} + ^{7}C_{3} + ^{7}C_{4} + ^{7}C_{5} + ^{7}C_{6} + ^{7}C_{7}

= 7 + \(\frac{7 \times 6}{2 \times 1}+\frac{7 \times 6 \times 5}{3 \times 2 \times 1}+\frac{7 \times 6 \times 5 \times 4}{4 \times 3 \times 2 \times 1}+\frac{7 \times 6 \times 5 \times 4 \times 3}{5 \times 4 \times 3 \times 2 \times 1}\) + 7 + 1

= 7 + 21 + 35 + 35 + 21 + 7 + 1

= 127

Question 14.

In how many ways can a hotel owner subscribe 3 newspapers and 2 magazines from 8 different newspapers and 5 different magazines available in the city? If a particular newspaper is to be selected and a particular magazine is not to be selected then in how many ways can this selection be done?

Answer:

8 different newspapers and 5 different magazines are available in the city.

A hotel owner can subscribe 3 newspapers out of 8 newspapers in ^{8}C_{3} ways and 2 magazines out of 5 magazines in ^{5}C_{2} ways

∴ Total combination = ^{8}C_{3} × ^{5}C_{2}

= \(\frac{8 \times 7 \times 6}{3 \times 2 \times 1}+\frac{5 \times 4}{2 \times 1}\)

= 56 × 10 = 560

If a particular newspaper is to be selected then 2 newspapers out of the remaining 7 newspapers can be subscribed in ^{7}C_{2} ways and a particular magazine is not to be selected then 2 magazines out of the remaining 4 magazines can be selected in ^{4}C_{2} ways.

∴ Total combinations = ^{7}C_{2} × ^{4}C_{2}

= \(\frac{7 \times 6}{2 \times 1} \times \frac{4 \times 3}{2 \times 1}\)

= 21 × 6

= 126

Question 15.

If ^{n}P_{2} + ^{n}C_{2} = 84 then find the value of n.

Answer:

^{n}P_{2} + ^{n}C_{2} = 84

∴ n(n – 1) = 84 × \(\frac{2}{3}\)

∴ n(n – 1) = 28 × 2

∴ n(n – 1) = 56

∴ n(n – 1) =8 × 7

∴ n(n – 1) = 8(8 – 1)

∴ n = 8

Question 16.

If ^{n}P_{r} ÷ ^{n}C_{r} = 24 then find the value of r.

Answer:

\(\frac{n \mathrm{P}_{r}}{n \mathrm{C}_{r}}\) = 24

∴ \(\frac{\frac{n !}{(n-r) !}}{\frac{n !}{r !(n-r) !}}\) = 24

∴ r! = 24

∴ r! = 4 × 3 × 2 × 1

∴ r! = 4!

∴ r = 4