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

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

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

Permutations and Combinations Exercise 6e Solutions

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

I.

Question 1.
A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of:
(i) Exactly 3 girls?
(ii) atleast 3 girls?
(iii) at most 3 girls?
Solution:
(i) Given exactly 3 girls
The total number of girls is 4.
Out of which 3 are to be choosen
∴ The number of ways in which we can choose would be made = ⁴C₃
The number of boys is 9, out of which 4 are to be choosen, which is given by ⁹C₄.
Total ways of forming the committee with exactly three girls.

(ii) Given atleast 3 girls
There are two possibilities for making a committee choosing atleast 3 girls.
Choosing three girls, we have done in (i) choosing four girls and 3 boys would be done in ⁴C₄ ways.
Total ways = ⁴C₄ × ⁹C₃

(iii) Given most 3 girls
0 girls and 7 boys
1 girl and 6 boys
2 girls and 5 boys
3 girls and 4 boys

The total number of ways in which a committee can have at most 3 girls is 36 + 336 + 756 + 504 = 1632

Question 2.
It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible?
Solution:
Given there is a total of 9 People
4 women can sit in four places, and the ways they can be seated = ⁴P₄
= \(\frac{4!}{(4-4)!}\)
= \(\frac{4 \times 3 \times 2 \times 1}{0!}\)
= 24
The number of ways in which these can be seated = ⁵P₅
= \(\frac{5!}{(5-5)!}\)
= \(\frac{5 \times 4 \times 3 \times 2 \times 1}{1}\)
24 × 120 = 2880

Question 3.
In how many ways can the letters of the word ASSASSINATION be arranged so that all the S’s are together?
Solution:
In the given word ASSASSINATION, there are 4 S’ since the 4 S’ have to be arranged together, let us take them.
The remaining letters are = 3A (2I, 2N, 1T)
Using the formula \(\frac{n!}{P_{1}!P_{2}!P_{3}!}\) when n is the number of terms and P₁P₂P₃ are the number of times the repeating letters repeat themselves.
Here P₁ = 3, P₂ = 2, P₃ = 2
Putting the values in the formula, we get
\(\frac{10!}{3!2!2!}=\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3!}{3!\times 2 \times 2 \times 1 \times 1}\) = 1512000

II.

Question 1.
How many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER?
Solution:
The word DAUGHTER has 3 vowels: A, E, and U, and 5 consonants: D, G, H, T, and R.
The three vowels can be choosen in ³C₂ as only two vowels are to be choosen.
Similarly, the five consonants can be choosen ⁵C₃ ways.
= \(\frac{3!}{2!(3-2)!} \times \frac{5!}{3!(5-3)!}\)
= \(\frac{3!}{2!1!} \times \frac{5!}{3!2!}\)
= 30
The total number of ways is 30.
\(\frac{5!}{(5-5)!}=\frac{5!}{0!}=\frac{5!}{1}\)
= 5 × 4 × 3 × 2 × 1
= 120

Question 2.
How many words, with or without meaning, can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together?
Solution:
In the word EQUATION, there are 5 vowels (A, E, I, O, U) and 3 consonants (Q, T, N).
The number of ways in which 5 vowels can be arranged is ⁵C₃.
= \(\frac{5!}{(5-5)!}\)
= \(\frac{5 \times 4 \times 3 \times 2 \times 1}{0!}\)
= \(\frac {120}{1}\)
= 120
Similarly, the number of ways in which 3 consonants can be arranged is ³C₃.
= \(\frac{3!}{(3-3)!}\)
= \(\frac{3 \times 2 \times 1}{0!}\)
= \(\frac {6}{1}\)
= 6
These are two ways in which vowels and consonants can appear together.
(AEIOU) (QTN) or (QTN) (AEIOU)
∴ 2 × 120 × 6 = 1440

Question 3.
If the different permutations of all the letters of the word EXAMINATION are listed as in a dictionary, how many words are there in this list before the first word starting with E?
Solution:
In a dictionary, words are listed alphabetically,
So to find the words listed before E should start with the letter either A, B, C, or D.
But the word EXAMINATION doesn’t have B, C, or D.
\(\frac{n!}{P_{1}!P_{2}!P_{3}!}\) = \(\frac{10!}{2!2!}\) = 907200

Question 4.
How many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7, and 9, which are divisible by 10 and no digit is repeated?
Solution:
The number is divisible by 10 in the unit place.
The 6-digit number is not to be formed out of which unit place is fixed as 0.
Here n = 5
And the number of choices available is 5.
So, the total ways in which the rest of the places can be filled is ⁵P₅.
= \(\frac{5!}{(5-5)!} \times 1 \frac{5!}{1} \times 1\)
= 5 × 4 × 3 × 2 × 1 × 1
= 120

Question 5.
The English alphabet has 5 vowels and 21 consonants. How many words with two different vowels and 2 different consonants can be formed from the alphabet?
Solution:
We know that there are 5 vowels and 21 consonants in the English alphabet.
Choosing two vowels out of 5 would be done in ⁵C₂ ways.
Choosing 2 consonants out of 21 can be done in ²¹C₂ ways.
The total number of ways to select 2 vowels and 2 consonants.

III.

Question 1.
In an examination, a question paper consists of 12 questions divided into two parts, i.e., Part I and Part II, containing 5 and 7 questions, respectively. A student is required to attempt 8 questions in all, selecting at least 3 from each part. In how many ways can a student select the questions?
Solution:
The student can choose 3 questions from Part I and 5 from Part II
3 questions from Part I and 5 from Part II can be choosen in ⁵C₃ × ⁷C₅ ways.
= \(\frac{5!}{3!2!} \times \frac{7!}{5!2!}\)
= \(\frac{5 \times 4 \times 3!}{3!\times 2 \times 1} \times \frac{7 \times 6 \times 5!}{5!\times 2 \times 1}\)
= 210
4 questions from Part-I and 4 from Part-II can be choosen in ⁵C₄ × ⁷C₄ ways.
= \(\frac{5!}{4!1!} \times \frac{7!}{4!3!}\)
= \(\frac{5 \times 4!}{4!} \times \frac{7 \times 6 \times 5 \times 4!}{4!\times 3 \times 2 \times 1}\)
= 175
5 Questions from Part I and 3 from Part II can be choosen in ⁵C₅ × ⁷C₃
= \(\frac{5!}{5!0!} \times \frac{7!}{3!4!}\)
= 1 × \(\frac{7 \times 6 \times 5 \times 4!}{3 \times 2 \times 1 \times 4!}\)
= 35
Now the total number of ways in which a student can choose the questions is 210 + 175 + 35 = 420.

Question 2.
Determine the number of 5-card combinations out of a deck of 52 cards if each selection of 5 cards has exactly one king.
Solution:
We have a deck of cards that has 4 kings.
The number of remaining cards is 52.
Ways of selecting a king from the deck = ⁴C₁
= \(\frac{4!}{1!3!} \times \frac{48!}{4!44!}\)
= \(\frac{4 \times 3!}{3!} \times \frac{48 \times 47 \times 46 \times 45 \times 44!}{4 \times 3 \times 2 \times 1 \times 44!}\)
= 778320

Question 3.
From a class of 25 students, 10 are to be choosen for an excursion party. 3 students decide that either all of them will join or none of them will join. In how many ways can the excursion party be chosen?
Solution:
In this question, we get 2 options, which are
(i) Either all 3 will go
Then, the remaining students in the class are 25 – 3 = 22.
The number of students to be chosen for the Party = 7
Number of ways to choose the remaining 22 students = ²²C₇
= \(\frac{22!}{7!15!}\)
= 170544

(ii) None of them will go
The students going will be 10
Remaining students eligible for going = 22
\(\frac{22!}{10!12!}\) = 646646
The Total number of ways in which students can be chosen is 170544 + 646646 = 817190.