# GSEB Solutions Class 11 Statistics Chapter 6 Permutations, Combinations and Binomial Expansion Ex 6.2

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) 11C4
nCr = $$\frac{n !}{r !(n-r) !}$$
11C4 = $$\frac{11 !}{41(11-4) !}$$
= $$\frac{11 \times 10 \times 9 \times 8 \times 7 !}{4 \times 3 \times 2 \times 1 \times 7 !}$$
= $$\frac{7920}{24}$$ = 330

(2) 9C0
nC0 = 1
9C0 = 1

(3) 25C23
nCr = $$\frac{n !}{r !(n-r) !}$$
25C23 = $$\frac{25 !}{23 !(25-23) !}$$
= $$\frac{25 !}{23 ! 2 !}$$
= $$\frac{25 \times 24 \times 23 !}{23 ! \times 2 \times 1}=\frac{600}{2}$$ = 300

(4) 8C8
nCr = 1
8C8 = 1 Question 2.
Find the unknown value:
(1) nC2 = 28
nC2 = 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) 27Cr+4 = 27C2r-1
27Cr+4 = 27C2r-1
nCx = nCy
Or
n = x + y
27Cr+4 = 27C2r-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) nCn-2 = 15
∴ $$\frac{n !}{(n-2) !(n-n+2) !}$$ = 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.nC4 = 7.nC3 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?
Total numbcr of combinations selecting 2 candidates for the post of peon out of 8 candidates
= 8C2
= $$\frac{8 !}{2 !(8-2) !}$$
= $$\frac{8 !}{2 ! 6 !}$$
= $$\frac{8 \times 7 \times 6 !}{2 \times 6 !}$$
=$$\frac{8 \times 7}{2}$$ = 28

Alternative method:
8C2 = $$\frac{8 \times 7}{2 \times 1}$$
= 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?
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 = 5C2
= $$\frac{5 \times 4}{2 \times 1}$$
= 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?
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 = 10C2
= $$\frac{10 \times 9 \times 8}{3 \times 2 \times 1}$$
= $$\frac{720}{6}$$ = 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?
3 clerks out of 14 clerks can be selected in 14C3 ways and one peon out of 6 peons can be selected in 6C1 ways.
∴ Total combinations = 14C3 × 6C1
= $$\frac{14 \times 13 \times 12}{3 \times 2 \times 1}$$ × 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?
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 3C3 ways or
3 pink flowers of 6 pInk flowers can be selected in 6C3 ways.
∴ Total combinations = 3C3 + 5C3
= 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 = 1C1 × 8C3
= 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?
(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 = 13C2 × 6C3
= $$\frac{13 \times 12}{2}$$
= 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 4C1 ways and one card of queen out of 4 cards of queen can be selected In 4C1 ways.
∴ Total combinations = 4C1 × 4C1
= 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?
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 8C3 ways.
∴ Total combinations = 1C1 × 8C3
= 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 in6C3 ways.
∴ Total combinations = 1C1 × 6C3
= 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?
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?
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
= 6C1 + 6C2 + 6C3 + 6C4 + 6C5 + 6C6
= 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?
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 7C4 ways.
∴ Total combinations = 1C1 × 7C4
= 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 = 7C5
= $$\frac{7 \times 6 \times 5 \times 4 \times 3}{5 \times 4 \times 3 \times 2 \times 1}$$
= 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?
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
= 7C1 + 7C2 + 7C3 + 7C4 + 7C5 + 7C6 + 7C7
= 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?
8 different newspapers and 5 different magazines are available in the city.
A hotel owner can subscribe 3 newspapers out of 8 newspapers in 8C3 ways and 2 magazines out of 5 magazines in 5C2 ways
∴ Total combination = 8C3 × 5C2
= $$\frac{8 \times 7 \times 6}{3 \times 2 \times 1}+\frac{5 \times 4}{2 \times 1}$$
= 56 × 10 = 560

If a particular newspaper is to be selected then 2 newspapers out of the remaining 7 newspapers can be subscribed in 7C2 ways and a particular magazine is not to be selected then 2 magazines out of the remaining 4 magazines can be selected in 4C2 ways.
∴ Total combinations = 7C2 × 4C2
= $$\frac{7 \times 6}{2 \times 1} \times \frac{4 \times 3}{2 \times 1}$$
= 21 × 6
= 126

Question 15.
If nP2 + nC2 = 84 then find the value of n.
nP2 + nC2 = 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 nPr ÷ nCr = 24 then find the value of r.
$$\frac{n \mathrm{P}_{r}}{n \mathrm{C}_{r}}$$ = 24
∴ $$\frac{\frac{n !}{(n-r) !}}{\frac{n !}{r !(n-r) !}}$$ = 24