Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Permutations and Combinations Solutions Exercise 5(e) will help students to clear their doubts quickly.

## Intermediate 2nd Year Maths 2A Permutations and Combinations Solutions Exercise 5(e)

I.

Question 1.
If nC4 = 210, find n.
Solution:
nCr = $$\frac{n !}{(n-r) ! r !}$$ = $$\frac{n(n-1)(n-2) \ldots \ldots[n-(r-1)]}{1.2 .3 \ldots \ldots \ldots . . . r}$$
Solution:
nC4 = 210
⇒ $$\frac{n(n-1)(n-2)(n-3)}{1.2 .3 .4}=10 \times 21^n C_{2 r-1}$$
⇒ n(n – 1) (n – 2) (n – 3) = 10 × 21 × 1 × 2 × 3 × 4
⇒ n(n – 1) (n – 2) (n – 3) = 10 × 7 × 3 × 2 × 3 × 4
⇒ n(n – 1) (n – 2) (n – 3) = 10 × 9 × 8 × 7
⇒ n = 10

Question 2.
If 12Cr = 495, find the possible values of r.
Solution:
Hint: nCr = nCn-r
12Cr = 495
= 5 × 99
= 11 × 9 × 5
= $$\frac{12 \times 11 \times 9 \times 5 \times 2}{12 \times 2}$$
= $$\frac{12 \times 11 \times 10 \times 9}{1.2 .3 .4}$$
= 12C4 or 12C8
∴ r = 4 or 8

Question 3.
If 10 . nC2 = 3 . n+1C3, find n.
Solution:
10 . nC2 = 3 . n+1C3
⇒ 10 × $$\frac{n(n-1)}{1.2}=\frac{3(n+1)(n-1)}{1.2 .3}$$
⇒ 10 = n + 1
⇒ n = 9

Question 4.
If nPr = 5040 and nCr = 210, find n and r.
Solution:
Hint: nPr = r! nCr and nPr = n(n – 1) (n – 2) ……. [n – (r – 1)]
nPr = 5040, nCr = 210
r! = $$\frac{{ }^n P_r}{{ }^n C_r}=\frac{5040}{210}=\frac{504}{21}$$ = 24 = 4!
∴ r = 4
nPr = 5040
nP4 = 5040
= 10 × 504
= 10 × 9 × 56
= 10 × 9 × 8 × 7
= 10P4
∴ n = 10
∴ n = 10, r = 4

Question 5.
If nC4 = nC6, find n.
Solution:
nCr = nCs ⇒ r = s or r + s = n
nC4 = nC6
∴ n = 4 + 6 = 10, (∵ 4 ≠ 6)

Question 6.
If 15C2r-1 = 15C2r+4, find r.
Solution:
15C2r-1 = 15C2r+4
2r – 1 = 2r + 4 or (2r – 1) + (2r + 4) = 15
(2r – 1) + (2r + 4) = 15
⇒ 4r + 3 = 15
⇒ 4r = 12
⇒ r = 3
∴ 2r – 1 = 2r + 4
⇒ -1 = 4 which is impossible
∴ r = 3

Question 7.
If 17C2t+1 = 17C3t-5, find t.
Solution:
17C2t+1 = 17C3t-5
2t + 1 = 3t – 5 or (2t + 1) + (3t – 5) = 17
⇒ 1 + 5 = t or 5t = 21
⇒ t = 6 or t = $$\frac{21}{5}$$ which is not an integer
∴ t = 6

Question 8.
If 12Cr+1 = 12C3r-5, find r.
Solution:
12Cr+1 = 12C3r-5
⇒ r + 1 = 3r – 5 or (r + 1) + (3r – 5) = 12
⇒ 1 + 5 = 2r or 4r – 4 = 12
⇒ 2r = 6 or 4r = 16
⇒ r = 3 or r = 4
∴ r = 3 or 4

Question 9.
If 9C3 + 9C5 = 10Cr then find r.
Solution:
nCr = nCn-r
10Cr = 9C3 + 9C5
9C3 + 9C5 = 9C3 + 9C5 = 10C6 or 10C4 = 10Cr (given)
⇒ r = 4 or 6

Question 10.
Find the number of ways of forming a committee of 5 members from 6 men and 3 ladies.
Solution:
Total number of persons = 6 + 3 = 9
∴ Number of ways of forming a committee of 5 members from 6 men and 3 ladies = 9C5
= $$\frac{9 \times 8 \times 7 \times 6 \times 5}{5 \times 4 \times 3 \times 2 \times 1}$$
= 126

Question 11.
In question no. 10, how many committees contain atleast two ladies?
Solution:
Since a committee contains atleast 2 ladies, the members of the committee may be of the following two types.
The number of selections in the first type = 6C3 × 3C2
= 20 × 3
= 60
The number of selections in the second type = 6C2 × 3C3
= 15 × 1
= 15
∴ The required number of ways of selecting the committee containing atleast 2 ladies = 60 + 15 = 75.

Question 12.
If nC5 = nC6, then find 13Cn.
Solution:
nC5 = nC6
⇒ n = 6 + 5 = 11
13Cn = 13C11 = 13C2
= $$\frac{13 \times 12}{1 \times 2}$$
= 78

II.

Question 1.
Prove that for 3 ≤ r ≤ n, (n-3)Cr . (n-3)C(r-1) + 3 . (n-3)C(r-2) + 3 . (n-3)C(r-3) = nCr
Solution:

Question 2.
Find the value of 10C5 + 2 . 10C4 + 10C3
Solution:
Hint: nCr + nCr-1 = (n+1)Cr

Question 3.
Simplify 34C5 + $$\sum_{r=0}^4{ }^{(38-r)} C_4$$
Solution:

Question 4.
In a class, there are 30 students. If each student plays a chess game with each of the other students then find the total number of chess games played by them.
Solution:
Number of students in a class = 30
Since each student plays a chess game with each of the other students, the total number of chess games played by them = 30C2 = 435

Question 5.
Find the number of ways of selecting 3 girls and 3 boys out of 7 girls and 6 boys.
Solution:
The number of ways of selecting 3 girls and 3 boys Out of 7 girls and 6 boys = 7C3 × 6C3
= 35 × 20
= 700

Question 6.
Find the number of ways of selecting a committee of 6 members out of 10 members always including a specified member.
Solution:
Since a specified member is always included in a committee, the remaining 5 members can be selected from the remaining 9 members in 9C5 ways.
∴ Required number of ways selecting a committee = 9C5 = 126

Question 7.
Find the number of ways of selecting 5 books from 9 different mathematics books such that a particular book is not included.
Solution:
Since a particular book is not included in the selection, the 5 books can be selected from the remaining 8 books in 8C5 ways.
∴ The required number of ways of selecting 5 books = 8C5 = 56

Question 8.
Find the number of ways of selecting 3 vowels and 2 consonants from the letters of the word EQUATION.
Solution:
The word EQUATION contains 5 vowels and 3 consonants.
The 3 vowels can be selected from 5 vowels in 5C3 = 10 ways.
The 2 consonants can be selected from 3 consonants in 3C2 = 3 ways.
∴ The required number of ways of selecting 3 vowels and 2 consonants = 10 × 3 = 30

Question 9.
Find the number of diagonals of a polygon with 12 sides.
Solution:
The number of diagonals of a polygon with sides = $$\frac{n(n-3)}{2}$$
= $$\frac{12(12-3)}{2}$$
= 54

Question 10.
If n persons are sitting in a row, find the number of ways of selecting two persons, who are sitting adjacent to each other.
Solution:
The number of ways of selecting 2 persons out of n persons sitting in a row, who are sitting adjacent to each other = n – 1

Question 11.
Find the number of ways of giving away 4 similar coins to 5 boys if each boy can be given any number (less than or equal to 4) of coins.
Solution:
The 4 similar coins can be divided into different groups as follows.
(i) One group containing 4 coins
(ii) Two groups containing 1, 3 coins respectively
(iii) Two groups containing 2, 2 coins respectively
(iv) Two groups containing 3, 1 coins respectively
(v) Three groups containing 1, 1, 2 coins respectively
(vi) Three groups containing 1, 2, 1 coins respectively
(vii) Three groups containing 2, 1, 1 coins respectively
(viii) Four groups containing 1, 1, 1, 1 coins respectively
these groups can given away to 5 boys in = $${ }^5 C_1+2 \times{ }^5 C_2+{ }^5 C_2+{ }^5 C_3 \times \frac{3 !}{2 !}+{ }^5 C_4$$
= 5 + 20 + 10 + 30 + 5
= 70 ways

III.

Question 1.
Prove that $$\frac{{ }^{4 n} C_{2 n}}{{ }^{2 n} C_n}=\frac{1.3 .5 \ldots \ldots(4 n-1)}{\{1.3 .5 \ldots \ldots(2 n-1)\}^2}$$
Solution:

Question 2.
If a set A has 12 elements, find the number of subsets of A having
(i) 4 elements
(ii) Atleast 3 elements
(iii) Atmost 3 elements
Solution:
Number of elements in set A = 12
(i) Number of subsets of A with exactly 4 elements = 12C4 = 495

(ii) The required subset contains atleast 3 elements.
The number of subsets of A with exactly 0 elements is 12C0
The number of subsets of A with exactly 1 element is 12C1
The number of subsets of A with exactly 2 elements is 12C2
Total number of subsets of A formed = 212
∴ Number of subsets of A with atleast 3 elements = (Total number of subsets) – (number of subsets contains 0 or 1 or 2 elements)
= 212 – (12C0 + 12C1 + 12C2)
= 4096 – (1 + 12 + 66)
= 4096 – 79
= 4017

(iii) The required subset contains atmost 3 elements
i.e., it may contain 0 or 1 or 2 or 3 elements.
The number of subsets of A with exactly 0 elements is 12C0
The number of subsets of A with exactly 1 element is 12C1
The number of subsets of A with exactly 2 elements is 12C2
The number of subsets of A with exactly 3 elements is 12C3
∴ Number of subsets of A with atmost 3 elements = 12C0 + 12C1 + 12C2 + 12C3
= 1 + 12 + 66 + 220
= 299

Question 3.
Find the number of ways of selecting a cricket team of 11 players from 7 batsmen and 6 bowlers such that there will be atleast 5 bowlers in the team.
Solution:
Since the team consists of at least 5 bowlers, the selection may be of the following types.

The number of selections in the first type = 7C6 × 6C5
= 7 × 6
= 42
The number of selections in the second type = 7C5 × 6C6
= 21 × 1
= 21
∴ The required number of ways selecting the cricket team = 42 + 21 = 63

Question 4.
If 5 vowels and 6 consonants are given, then how many 6-letter words can be formed with 3 vowels and 3 consonants?
Solution:
No. of vowels given = 5
No.of consonants given = 6
We have to form a 6-letter word with 3 vowels and 3 consonants from given letters.
3 vowels can select from 5 in 5C3 ways.
3 consonants can select from 6 in 6C3 ways.
Total No. of words = 5C3 × 6C3 × 6! = 1,44,000

Question 5.
There are 8 railway stations along a railway line. In how many ways can a train be stopped at 3 of these stations such that no two of them are consecutive?
Solution:
Number of ways of selecting 3 stations out of 8 = 8C3 = 56
Number of ways of selecting 3 out of 8 stations such that 3 are consecutive = 6
Number of ways of selecting 3 out of 8 stations such that 2 of them are consecutive = 2 × 5 + 5 × 4
= 10 + 20
= 30
∴ Number of ways for a train to be stopped at 3 of 8 stations such that no two of them are consecutive = 56 – (6 + 30) = 20

Question 6.
Find the number of ways of forming a committee of 5 members out of 6 Indians and 5 Americans so that always the Indians will be in majority in the committee.
Solution:
Since the committee contains the majority of Indians, the members of the committee may be of the following types.

The number of selections in type I = 6C5 × 5C0 = 6 × 1 = 6
The number of selections in type II = 6C4 × 5C1 = 15 × 5 = 75
The number of selections in type III = 6C3 × 5C2 = 20 × 10 = 200
∴ The required number ways of selecting a committee = 6 + 75 + 200 = 281.

Question 7.
A question paper is divided into 3 sections A, B, C Containing 3, 4, 5 questions respectively. Find the number of ways of attempting 6 questions choosing at least one from each section.
Solution:
First Method: The selection of a question may be of the following

Total No. of ways of attempting 6 questions

Second Method:
Required No.of attempting 6 questions = Total no. of arrangements – selection except question from C – selection except Q from A – selection except Q from B
= 12C67C69C66C6
= 805

Question 8.
Find the number of ways in which 12 things be
(i) divided into 4 equal groups
(ii) distributed to 4 persons equally.
Solution:
(i) The number of ways in which 12 things be divided into 4 equal groups = $$\frac{12 !}{3 ! 3 ! 3 ! 3 ! 4 !}$$ = $$\frac{12 !}{(3 !)^4 4 !}$$
(ii) The number of ways in which 12 things be distributed to 4 persons equally = $$\frac{12 !}{3 ! 3 ! 3 ! 3 !}$$ = $$\frac{12 !}{(3 !)^4}$$

Question 9.
A class contains 4 boys and g girls. Every Sunday, five students with atleast 3boys go for a picnic. A different group is being sent every week. During the picnic, the class teacher gives each girl in the group a doll. If the total number of dolls distributed is 85, find g.
Solution:
No. of boys = 4
No. of girls = g
Since there should be atleast 3 boys it can be done in 2 ways as shown in the table

The number of girls in G1 = [4C3 × gC2] × 2
Since each group contains 2 girls
The number of girls in G2 = [4C3 × gC2] × 1
Since each group contains 1 girl.
Given no. of dolls distributed = 85
⇒ [4C3 × gC2] × 2 + [4C4 × gC1] × 1 = 85
⇒ 4 . $$\frac{g(g-1)}{2}$$ × 2 + 1 . g . 1 = 85
⇒ 4g2 – 4g + g – 85 = 0
⇒ 4g2 – 3g – 85 = 0
⇒ 4g2 – 20g + 17g – 85 = 0
⇒ 4g(g – 5) + 17(g – 5) = 0
⇒ (g – 5)(4g + 17) = 0
Since g ≠ $$\frac{-17}{4}$$
∴ g = 5
Hence No. of girls = 5