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 ^{n}C_{4} = 210, find n.

Solution:

^{n}C_{r} = \(\frac{n !}{(n-r) ! r !}\) = \(\frac{n(n-1)(n-2) \ldots \ldots[n-(r-1)]}{1.2 .3 \ldots \ldots \ldots . . . r}\)

Solution:

^{n}C_{4} = 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 ^{12}C_{r} = 495, find the possible values of r.

Solution:

Hint: ^{n}C_{r} = ^{n}C_{n-r}

^{12}C_{r} = 495

= 5 × 99

= 11 × 9 × 5

= \(\frac{12 \times 11 \times 9 \times 5 \times 2}{12 \times 2}\)

= \(\frac{12 \times 11 \times 10 \times 9}{1.2 .3 .4}\)

= ^{12}C_{4} or ^{12}C_{8}

∴ r = 4 or 8

Question 3.

If 10 . ^{n}C_{2} = 3 . ^{n+1}C_{3}, find n.

Solution:

10 . ^{n}C_{2} = 3 . ^{n+1}C_{3}

⇒ 10 × \(\frac{n(n-1)}{1.2}=\frac{3(n+1)(n-1)}{1.2 .3}\)

⇒ 10 = n + 1

⇒ n = 9

Question 4.

If ^{n}P_{r} = 5040 and ^{n}C_{r} = 210, find n and r.

Solution:

Hint: ^{n}P_{r} = r! ^{n}C_{r} and ^{n}P_{r} = n(n – 1) (n – 2) ……. [n – (r – 1)]

^{n}P_{r} = 5040, ^{n}C_{r} = 210

r! = \(\frac{{ }^n P_r}{{ }^n C_r}=\frac{5040}{210}=\frac{504}{21}\) = 24 = 4!

∴ r = 4

^{n}P_{r} = 5040

^{n}P_{4} = 5040

= 10 × 504

= 10 × 9 × 56

= 10 × 9 × 8 × 7

= ^{10}P_{4}

∴ n = 10

∴ n = 10, r = 4

Question 5.

If ^{n}C_{4} = ^{n}C_{6}, find n.

Solution:

^{n}C_{r} = ^{n}C_{s} ⇒ r = s or r + s = n

^{n}C_{4} = ^{n}C_{6}

∴ n = 4 + 6 = 10, (∵ 4 ≠ 6)

Question 6.

If ^{15}C_{2r-1} = ^{15}C_{2r+4}, find r.

Solution:

^{15}C_{2r-1} = ^{15}C_{2r+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 ^{17}C_{2t+1} = ^{17}C_{3t-5}, find t.

Solution:

^{17}C_{2t+1} = ^{17}C_{3t-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 ^{12}C_{r+1} = ^{12}C_{3r-5}, find r.

Solution:

^{12}C_{r+1} = ^{12}C_{3r-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 ^{9}C_{3} + ^{9}C_{5} = ^{10}C_{r} then find r.

Solution:

^{n}C_{r} = ^{n}C_{n-r}

^{10}C_{r} = ^{9}C_{3} + ^{9}C_{5}

∴ ^{9}C_{3} + ^{9}C_{5} = ^{9}C_{3} + ^{9}C_{5} = ^{10}C_{6} or ^{10}C_{4} = ^{10}C_{r} (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 = ^{9}C_{5}

= \(\frac{9 \times 8 \times 7 \times 6 \times 5}{5 \times 4 \times 3 \times 2 \times 1}\)

= 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.

(i) 3 men, 2 ladies

(ii) 2 men, 3 ladies

The number of selections in the first type = ^{6}C_{3} × ^{3}C_{2}

= 20 × 3

= 60

The number of selections in the second type = ^{6}C_{2} × ^{3}C_{3}

= 15 × 1

= 15

∴ The required number of ways of selecting the committee containing atleast 2 ladies = 60 + 15 = 75.

Question 12.

If ^{n}C_{5} = ^{n}C_{6}, then find ^{13}C_{n}.

Solution:

∵ ^{n}C_{5} = ^{n}C_{6}

⇒ n = 6 + 5 = 11

^{13}C_{n} = ^{13}C_{11} = ^{13}C_{2}

= \(\frac{13 \times 12}{1 \times 2}\)

= 78

II.

Question 1.

Prove that for 3 ≤ r ≤ n, ^{(n-3)}C_{r} . ^{(n-3)}C_{(r-1)} + 3 . ^{(n-3)}C_{(r-2)} + 3 . ^{(n-3)}C_{(r-3)} = ^{n}C_{r}

Solution:

Question 2.

Find the value of ^{10}C_{5} + 2 . ^{10}C_{4} + ^{10}C_{3}

Solution:

Hint: ^{n}C_{r} + ^{n}C_{r-1} = ^{(n+1)}C_{r}

Question 3.

Simplify ^{34}C_{5} + \(\sum_{r=0}^4{ }^{(38-r)} C_4\)

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 = ^{30}C_{2} = 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 = ^{7}C_{3} × ^{6}C_{3}

= 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 ^{9}C_{5} ways.

∴ Required number of ways selecting a committee = ^{9}C_{5} = 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 ^{8}C_{5} ways.

∴ The required number of ways of selecting 5 books = ^{8}C_{5} = 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 ^{5}C_{3} = 10 ways.

The 2 consonants can be selected from 3 consonants in ^{3}C_{2} = 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 = ^{12}C_{4} = 495

(ii) The required subset contains atleast 3 elements.

The number of subsets of A with exactly 0 elements is ^{12}C_{0}

The number of subsets of A with exactly 1 element is ^{12}C_{1}

The number of subsets of A with exactly 2 elements is ^{12}C_{2}

Total number of subsets of A formed = 2^{12}

∴ Number of subsets of A with atleast 3 elements = (Total number of subsets) – (number of subsets contains 0 or 1 or 2 elements)

= 2^{12} – (^{12}C_{0} + ^{12}C_{1} + ^{12}C_{2})

= 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 ^{12}C_{0}

The number of subsets of A with exactly 1 element is ^{12}C_{1}

The number of subsets of A with exactly 2 elements is ^{12}C_{2}

The number of subsets of A with exactly 3 elements is ^{12}C_{3}

∴ Number of subsets of A with atmost 3 elements = ^{12}C_{0} + ^{12}C_{1} + ^{12}C_{2} + ^{12}C_{3}

= 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 = ^{7}C_{6} × ^{6}C_{5}

= 7 × 6

= 42

The number of selections in the second type = ^{7}C_{5} × ^{6}C_{6}

= 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 ^{5}C_{3} ways.

3 consonants can select from 6 in ^{6}C_{3} ways.

Total No. of words = ^{5}C_{3} × ^{6}C_{3} × 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 = ^{8}C_{3} = 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 = ^{6}C_{5} × ^{5}C_{0} = 6 × 1 = 6

The number of selections in type II = ^{6}C_{4} × ^{5}C_{1} = 15 × 5 = 75

The number of selections in type III = ^{6}C_{3} × ^{5}C_{2} = 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

= ^{12}C_{6} – ^{7}C_{6} – ^{9}C_{6} – ^{6}C_{6}

= 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 G_{1} = [^{4}C_{3} × ^{g}C_{2}] × 2

Since each group contains 2 girls

The number of girls in G_{2} = [^{4}C_{3} × ^{g}C_{2}] × 1

Since each group contains 1 girl.

Given no. of dolls distributed = 85

⇒ [^{4}C_{3} × ^{g}C_{2}] × 2 + [^{4}C_{4} × ^{g}C_{1}] × 1 = 85

⇒ 4 . \(\frac{g(g-1)}{2}\) × 2 + 1 . g . 1 = 85

⇒ 4g^{2} – 4g + g – 85 = 0

⇒ 4g^{2} – 3g – 85 = 0

⇒ 4g^{2} – 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