Inter 1st Year Maths Permutations and Combinations Solutions Exercise 6d

Practicing the AP Board Solutions Class 11 Maths and Chapter 6 Inter 1st Year Maths Permutations and Combinations Solutions Exercise 6d Pdf Download will help students to clear their doubts quickly.

Intermediate 1st Year Maths Permutations and Combinations Solutions Exercise 6d

Permutations and Combinations Exercise 6d Solutions

Permutations and Combinations Class 11 Exercise 6d Solutions – Permutations and Combinations 6d Exercise Solutions

I.

Question 1.
If ⁿC₈ = ⁿC₂, find ⁿC₂.
Solution:
Given ⁿC₈ = ⁿC₂
we know that if ⁿC_r = ⁿC_p then either r = p or n = P
Here ⁿC₈ = ⁿC₂
⇒ 8 = n – 2
On rearranging, we get
⇒ n = 10
Now, ⁿC₂ = ¹⁰C₂
= \(\frac{10!}{2!(10-2)!}\)
= \(\frac{10 \times 9 \times 8!}{2 \times 1 \times 8!}\)
= \(\frac {90}{2}\)
= 45

Question 2.
Determine n if
(i) ²ⁿC₃ : ⁿC₃ = 12 : 1
(ii) ²ⁿC₃ : ⁿC₃ = 11 : 1
Solution:
(i) Given ²ⁿC₃ : ⁿC₃ = 12 : 1
The above equation can be written as
\(\frac{{ }^{2 \mathrm{n}} C_2}{{ }^{\mathrm{n}} C_3}=\frac{12}{1}\)
Substituting the formula, we get

Simplifying and computing
⇒ 4 × (2n – 1) = 12 × (n – 2)
⇒ 8n – 4 = 12n – 24
⇒ 4n = 20
⇒ n = 5

(ii) Given ²ⁿC₃ : ⁿC₃ = 11 : 1

⇒ 4 × (2n – 1) = 11 × (n – 2)
⇒ 8n – 4 = 11n – 22
⇒ 11n – 8n = 22 – 4
⇒ 3n = 18
⇒ n = 6

Question 3.
How many chords can be drawn through 21 points on a circle?
Solution:
Given 21 points on a circle
We know that we require two points on the circle to draw a chord
The number of chords = ²¹C₂
= \(\frac{21!}{2!(21-2)!}\)
= \(\frac{21 \times 20 \times 19!}{2!\times 19!}\)
= 21 × 10
= 210
∴ The total number of chords that can be drawn is 210.

Question 4.
In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?
Solution:
Given 5 boys and 4 girls in total
We can select 3 boys from 5 boys in ⁵C₃ ways
Similarly, we can choose 3 boys from 4 girls in ⁴C₃ ways.

Question 5.
If ⁿC₄ = 210, find n.
Solution:
ⁿC₄ = 210
⇒ \(\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3)}{1 \cdot 2 \cdot 3 \cdot 4}\) = 10 × 21
⇒ 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 6.
If ¹²C_r = 495, find the possible values of ‘r’.
Solution:

Question 7.
If 10 . ⁿC₂ = 3 . ⁿ⁺¹C₃, find n.
Solution:
10 . ⁿC₂ = 3 . ⁿ⁺¹C₃
⇒ \(10 \times \frac{\mathrm{n}(\mathrm{n}-1)}{1.2}=\frac{3(\mathrm{n}+1) \mathrm{n}(\mathrm{n}-1)}{1.2 .3}\)
⇒ 10 = n + 1
⇒ n = 9

Question 8.
If ¹⁵C_2r-1 = ¹⁵C_2r+4, find r.
Solution:
¹⁵C_2r-1 = ¹⁵C_2r+4
⇒ 2r – 1 = 2r + 4 or (2r – 1) + (2r + 4) = 15
⇒ 4r + 3 = 15
⇒ 4r = 12
⇒ r = 3

Question 9.
If ¹²C_r+1 = ¹²C_3r-5, find r.
Solution:
¹²C_r+1 = ¹²C_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

Question 10.
If ⁿC₅ = ⁿC₆, then find ¹³C_n.
Solution:
ⁿC₅ = ⁿC₆
⇒ n = 5 + 6 = 11
∴ ¹³C_n = ¹³C₁₁ = ¹³C₂ = 78

II.

Question 1.
Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls, and 5 blue balls if each selection consists of 3 balls of each colour.
Solution:
Given 6 red balls, 5 white balls, and 5 blue balls.
We can select 3 red balls from 6 red balls in ⁶C₃ ways.
Similarly, we can select 3 white balls from 5 white balls in ⁵C₃ ways.
Similarly, we can select 3 blue balls in ⁵C₃ ways.
∴ The number of ways of selecting 9 balls

Question 2.
Determine the number of 5-card combinations out of a deck of 52 cards if there is exactly one ace in each combination.
Solution:
Given a deck of 52 cards.
There are 4 Ace cards in a deck of 52 cards.
According to the question, we need to select 1 Ace card out of the 4 Ace cards.
∴ The number of ways to choose 1 Ace from 4 Ace cards is ⁴C₁.

Question 3.
In how many ways can one select a cricket team of eleven 17 players, in which only 5 players can bowl, if each cricket team of 11 must include exactly 4 bowlers?
Solution:
Given 17 players, in which only 5 players can bowl, if each Cricket team of 11 must include exactly 4 bowlers.
5 players can bowl, and we can require 4 bowlers in a team of 11.
Since we need 11 players in a team and already 4 bowlers have been selected, we need to select 7 more players from 12.
The number of ways we can select these players is ¹²C₇
The number of ways we can select these players is ⁵C₄ × ¹²C₇

Question 4.
A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.
Solution:
Given a bag that contains 5 black and 6 red balls.
The number of ways we can select 2 black balls from 5 black balls is ⁵C₂.
The number of ways we can select 3 red balls from 6 red balls is ⁶C₃.
The number of ways 2 black and 3 red balls can be selected is ⁵C₂ × ⁶C₃

Question 5.
In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?
Solution:
Given that 9 causes are available, and 2 specific causes are compulsory for every student.
Here, 2 courses are compulsory out of 9 courses, so a student needs to select 5 – 2 = 3 courses

Question 6.
Prove that for 3 ≤ r ≤ n.
\({ }^{(n-3)} C_r+3^{(n-3)} C_{r-1}+3^{(n-3)} C_{r-2}+{ }^{(n-3)} C_{r-3}\) = ⁿC_r.
Solution:

Question 7.
Simplify \({ }^{34} C_5+\sum_{r=0}^4{ }^{(38-r)} C_4\).
Solution:

Question 8.
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 9.
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 10.
Find the number of diagonals of a polygon with 12 sides.
Solution:
Here n = 12
Number of diagonals of ‘n’ sided polygon = \(\frac{n(n-3)}{2}\)
= \(\frac{12 \times 9}{2}\)
= 54

Question 11.
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.

III.

Question 1.
Prove that \(\frac{{ }^{4 n} C_{2 n}}{{ }^{2 n} C_n}=\frac{1.3 .5 \ldots .(4 n-1)}{\{1.3 .5 \ldots . .(2 n-1)\}^2}\)
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) At most 3 elements
Solution:
The number of elements in set A is 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 containing 0 or 1 or 2 elements)
= 2¹² – (¹²C₀ + ¹²C₁ + ¹²C₂)
= 4096 – (1 + 12 + 66)
= 4096 – 79
= 4017

(iii) The required subset contains at most 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.

Question 3.
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:
Reduce the problem to placing 3 items in available spots with space between.
For 3 non-consecutive choices from 8 stations, we do the following
Total available spots = 8 – (3 – 1) = 6
Choose 3 stations from these 6 positions = ⁶C₃
= \(\frac{6 \times 5 \times 4}{3 \times 2 \times 1}\)
= 20
Given that each 5-card combination should contain 1 ace card and 4 non-ace cards.

Question 4.
A question paper is divided into 3 sections: A, B, and C, containing 3, 4, and 5 questions respectively. Find the number of ways of attempting 6 questions, choosing atleast one from each section.
Solution:
Section A: 3 Questions (choose a)
Section B: 4 Questions (choose b)
Section C: 5 Questions (choose c)

Hence, attempt 6 questions choosing atleast one from each Section is 60 + 180 + 120 + 15 + 120 + 180 + 60 + 40 + 30 = 805 ways.

Question 5.
A class contains 4 boys and g girls. Every Sunday, five students, including at least 3 boys, 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:
Step 1: Possible group compositions
Each group has 5 students and at least 3 boys:
3 boys + 2 girls
4 boys + 1 girl
(5 boys is not possible since there are only 4 boys)
Step 2: Count such groups
Case 1: 3 boys + 2 girls
Ways to choose 3 boys out of 4 = C(4, 3) = 4
Ways to choose 2 girls out of g = C(g, 2)
Total such groups = 4 × C(g, 2)
Case 2: 4 boys + 1 girl
Ways to choose 4 boys out of 4 = C(4, 4) = 1
Ways to choose 1 girl out of g = C(g, 1) = g
Total such groups = g
Step 3: Total number of dolls distributed
Each 3 boys + 2 girls group contributes 2 dolls = 4 × C(g, 2) × 2 = 8 × C(g, 2)
Each 4 boys + 1 girl group contributes 1 doll = g
Total dolls = 8 × C(g, 2) + g = 85
Using c(g, 2) = \(\frac{g(g-1)}{2}\)
⇒ 8 × \(\frac{g(g-1)}{2}\) = 85
⇒ 4g(g – 1) + g = 85
⇒ 4g² – 4g + g = 85
⇒ 4g² – 3g – 85 = 0
Step 4: Solve the quadratic:
Using the quadratic formula:
g = \(\frac{[3 \pm(\sqrt{9}+1360)]}{8}\)
= \(\frac{[3 \pm \sqrt{1369}]}{8}\)
= \(\frac{[3 \pm 37]}{8}\)
Valid solution: g = \(\frac{(3+37)}{8}\) = 5
The number of girls in the class is g = 5.