RBSE Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Miscellaneous exercise

Rajasthan Board RBSE Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Miscellaneous exercise Textbook Exercise Questions and Answers.

RBSE Class 11 Maths Solutions Chapter 7 Permutations and Combinations Miscellaneous exercise 

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?
Answer:
No. of letters in the word DAUGHTER = 8
No. of vowels in it = 3 (A, U, E)
No. of consonants = 5 (D, G, H, T, R)
Now, no. of ways to select 2 vowels out of 3 vowels = ³C₂
and no. of ways to select 3 consonants out of 5 consonants = ⁵C₃
Thus, no of combinations formed by 2 vowels and 3 consonants

since, there are 5 letters in combination in which no.of permutations by taking all = ⁵P₅
Thus, total no. of permutations of 30 combinations

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?
Answer:
In word EQUATION
No. of consonants = 3 (Q, T, N)
No. of vowels = 5 (E, U, A, L O)
When vowels and consonants remains together, then no. of ways to arrange them =2!
Similarly,
No. of ways to arrange 5 vowels = 5!
No. of ways to arrange 3 consonants = 3!
By fundamental theorem of calculation, no. of permutations
= 2! × 5! × 3!
= 2 × 1 × 5 × 4 × 3 × 2 × 1 × 3 × 2 × 1 = 1440
Thus, required no. of words = 1440

Question 3.
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) at least 3 girls?
(iii) atmost 3 girls?
Answer:
No. of boys = 9
No. of girls = 4
No. of members in committee = 7

(i) If no. of girls = 3
No. of boys = 7 - 3 = 4
Then we have select 4 boys out of 9 and 3 girls out of 4 girls in committee.
Thus, no. of ways to choose 3 girls out of 4 girls = ⁴C₃
No. of ways to choose 4 boys out of 9 boys = ⁹C₄
Then, total ways to form team of boys and 3 girls
= ⁴C₃ × ⁹C₄

= 4 × 9 × 7 × 2 = 504

(ii) If committee consist at least 3 girls
Hence, committee of 7 members can be formed in the following way:
(a) 3 Girls and 4 boys
(b) 4 girls and 3 boys ,
(a) If 3 girls and 4 boys remain in committee, then no. of ways to choose
= ⁴C₃ × ⁹C₄

= 4 × 126
= 504

(b) If 4 girls and 3 boys remain in committee then no. of ways t0 choose

Thus, total ways to form committee = 504 + 84 = 588

When committee consist of atmost 3 girls
Then commitee can be formed in the following way
Taking:
(a) 1 girl + 6 boys
(b) 2 girls + 5 boys
(c) 3 girls + 4 boys
(d) 0 girl or no girl, only 7 boys

(a) Case I
No. of ways, to form committee of 1 girl and 6 boys

= 4 × 84 = 336

(b) Case II
Number of ways to form committee of 2 girls and 5 boys

(c) Case III
Number of ways to form committee of 3 girls and 4 boys

(d) Case IV
If no girl in committee then all 7 will be boys Now, no. of ways to form committee

= 9 × 4 = 36
Now, total ways to form committee
= 336 + 756 + 504 + 36 = 1632
Thus, required no. of ways = 1632

Question 4.
If the different permutations of all the letter 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?
Answer:
No. of letters in word EXAMINATION = 11
Here, A occurs twice, I twice and N twice. According to alphabet first letter will be A and then E, etc as
A E I M N O T X
Always A will be the first letter before the first word starting with E
Now fix A and permutation of remaining 10 letters with A will formed words. In these 10 letters I occurs twice and M also twice.
Now, no. of permutations formed by these 10 letters 10!
= \(\frac{10 !}{2 ! 2 !}\)
= \(\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 1 \times 2 \times 1}\)
= 907200
Thus, no. of words are there in the list before the first word starting with E = 907200

Question 5.
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?
Answer:
No. of digits = 6 (0,1, 3, 5,7, 9)
Using these 6 numbers we have to form the numbers which are divisible by 10.
Numbers divisible by 10 have unit place 0.
Now fix 0 at unit place.
Numbers from remaining five digits 1, 3, 5, 7, 9 in which digits are not repeated. .
= ⁵p₅ = 5! [∵ ⁿP_n = n!]
= 5 × 4 × 3 × 2 × 1 = 120
Thus, required numbers = 120

Question 6.
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?
Answer:
No. of vowels in alphabet = 5
Then, no, of combinations formed by 2 different vowels

No. of consonants in alphabet = 21
No. of combinations formed by 2 distinct consonants

Thus, total combinations of 2 vowels and two consonants
= 210 × 10 = 2100
Now, no. of permutations formed by 2 vowels and 2 consonants i.e. 4 distinct letters
= ⁴P₄ = 4! = 4 × 3 × 2 × 1 = 24
By fundamental theorem of calculation,
No. of total permutations = 2100 × 24 = 50400
Thus, total no. of words formed by 2 vowels and 2 consonants of alphabet = 50400.

Question 7.
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?
Answer:
Given question paper is divided into two parts I and II part which contains 5 and 7 questions respectively.
A student has to select total 8 questions whereas he can select at least 3 questions from each part. Students can select questions in the following way.
(i) 3 questions from part I and 5 questions from part II
(ii) 4 questions from part I and 4 questions from part II
(iii) 5 questions from part I and 3 questions from part II

(i) In this student has to select 3 questions out of 5 questions and 5 questions out of 7 questions from part I and part II respectively.
Then, total possible ways of selection

(ii) In this, student has to select 4 questions from each part
Thus, no. of ways to choose

(iii) In this, student has choose 5 questions from part I and 3 questions from part II
Then, total no. of ways to choose

Thus, total no. of ways, student can se1tct 8 questions
= 210 + 175 + 35 = 420
Thus, student can choose questions by 420 ways

Question 8.
Determine the number of 5 card combinations out of a deck of 52 cards if each selection of 5 cards has exactly one king.
Answer:
No. of kings in a deck of 52 cards = 4
∵ We have select 5 cards and in selection 1 king in necessary which is selected out of 4 kings in a pack

Now, 4 cards has to be select from remaining (52 - 4) = 48 cards of pack.
Then, total ways = ⁴⁸C₄

Then, by fundamental theorem of calculation, no. of combinations of 5 cards
= 4 × 194580
= 778320
Thus, total 5 card combinations = 778320

Question 9.
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?
Answer:
5 men and 4 women can be seated in a row at 9 places in which 1, 3, 5, 7 and 9 places are odd and 2, 4, 6, 8 places are even
So arrange as follows :

Now, total ways 4 wometi seated at 4 even places
= ⁴P₄ [∵ ⁿp_n = n!]
= 4!
= 4 × 3 × 2 × 1 = 24
Then, total ways, 5 men seated at 5 odd places
= ⁵P₅
= 5!
= 5 × 4 × 3 × 2 × 1 = 120
Then, 4 women and 5 men can seated in a row by
24 × 120 = 2880 ways
Thus, required way of sitting = 2880

Question 10.
From a class of 25 students, 10 are to be chosen for an excursion party. There are 3 students who decide that either all of them will join or none of them will join. In how many ways can the excursion party be chosen?
Answer:
Total no. of students in a class = 25
No. of students chosen for an excursion party = 10
According to question,
There are 3 students who decide that either all of them will join or none of them will join.

(i) If all 3 students join excursion party then out of remaining (25 - 3) = 22 students, (10 - 3) = 7 students will be selected
Then, no. of ways to choose 7 students out of 22

= 170544

(ii) If all 3 students did not join excursion party then we will select 10 students out of (25 - 3) = 22 students
Then, no, of ways to choose

= 646646
Thus, total ways of selection = 170544 + 646646 = 817190
Thus, the excursion party can be chosen by 817190 ways.

Question 11.
In how many ways can the letters of the word ASSASSINATION be arranged so that all the S’s are together?
Answer:
Total Letters in word ASSASS1NATLON 13
Here, A occurs 3 times
S occurs 4 times
J occurs 2 times
N occurs 2 times
and remaining letters are distinct.
Now all 4 S’s are together, so assumed them as 1 letter.
No. of permutations formed by 10 letters

Thus, required no.of ways = 151200