Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 7 Permutations and Combinations Miscellaneous Exercise Questions and Answers.
PSEB Solutions for Class 11 Maths 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.
There are 8 letters in the word ‘DAUGHTER’ including 3 vowels and 5 consonants.
We have to select 2 vowels out of 3 vowels and 3 consonants out of 5 consonants.
∴ Number of ways of selection = \({ }^{3} C_{2} \times{ }^{5} C_{3}\) = 3 × 10 = 30
Now, each word contains 5 letters which can be arranged among themselves in 5! ways.
So, total number of words = 5 ! x 30 = 120 × 30 = 3600
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.
The word EQUATION consists of 5 vowels and 3 consonants.
∴ 5 vowels can be arranged in 5! = 120 ways.
3 consonants can be arranged in 3! = 6 ways
The two blockof vowels and consonants can be arranged in 2! = 2 ways.
∴ The no. of world which can be formed with 1 letters of the word EQUATION so that vowels and consonants occur together = 120 × 6 × 2 = 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) atleast 3 girls?
(iii) atmost 3 girls?
Answer.
A committee of 7 has to be formed from 9 boys and 4 girls.
(i) Since exactly 3 girls are to be there in every committee, each committee must consist of (7 – 3) = 4 boys only.
Thus, in this case, required number of ways = \({ }^{4} C_{3} \times{ }^{9} C_{4}\)
= \(\frac{4 !}{3 ! 1 !} \times \frac{9 !}{4 ! 5 !}\)
= \(4 \times \frac{9 \times 8 \times 7 \times 6 \times 5 !}{4 \times 3 \times 2 \times 1 \times 5 !}\) = 504.
(ii) Since atleast 3 girls are to be there in every committee, the committee can consist of
(a) 3 girls and 4 boys or
(b) 4 girls and 3 boys
3 girls and 4 boys can be selected in \({ }^{4} C_{3} \times{ }^{9} C_{4}\) ways.
4 girls and 3 boys can be selected in \({ }^{4} \mathrm{C}_{4} \times{ }^{9} \mathrm{C}_{3}\) ways.
Therefore, in this case, required number of ways
= \({ }^{4} C_{3} \times{ }^{9} C_{4}\) + \({ }^{4} \mathrm{C}_{4} \times{ }^{9} \mathrm{C}_{3}\)
= 504 + 84 = 588.
(iii) Since atmost 3 girls are to be there in every committee, the committee can consist of
(a) 3 girls and 4 boys
(b) 2 girls and 5 boys
(c) 1 girl and 6 boys
(d) No girl and 7 boys
3 girls and 4 boys can be selected in \({ }^{4} C_{3} \times{ }^{9} C_{4}\) ways.
2 girls and 5 boys can be selected in \({ }^{4} C_{2} \times{ }^{9} C_{5}\) ways.
1 girl and 6 boys can be selected in \({ }^{4} C_{1} \times{ }^{9} C_{6}\) ways.
No girl and 7 boys can be selected in \({ }^{4} C_{0} \times{ }^{9} C_{7}\) ways.
Therefore, in this case, required number, of ways :
= \({ }^{4} \mathrm{C}_{3} \times{ }^{9} \mathrm{C}_{4}+{ }^{4} \mathrm{C}_{2} \times{ }^{9} \mathrm{C}_{5}+{ }^{4} \mathrm{C}_{1} \times{ }^{9} \mathrm{C}_{6}+{ }^{4} \mathrm{C}_{0} \times{ }^{9} \mathrm{C}_{7}\)
= \(\frac{4 !}{3 ! 1 !} \times \frac{9 !}{4 ! 5 !}+\frac{4 !}{2 ! 2 !} \times \frac{9 !}{5 ! 4 !}+\frac{4 !}{1 ! 3 !} \times \frac{9 !}{6 ! 3 !}+\frac{4 !}{0 ! 4 !} \times \frac{9 !}{7 ! 2 !}\)
= 504 + 756 + 336 + 36 = 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.
Words starting with A are formed with the letters 2I’s, 2N’s, A, E, X, M, T, O.
Number of words formed by these letters = \(\frac{10 !}{2 ! 2 !}\)
= \(\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 2}\) = 907200
Then the words starting with E, I, M, N, O, T, X will be formed.
∴ Number of words before the first word starting with E is formed = 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.
A number is divisible by 10 if its units digits is 0.
Therefore, 0 is fixed at the units place.
Therefore, there will be as many ways as there are ways of filling 5 vacant places 
in succession by the remaining 5 digits (i.e., 1, 3, 5, 7 and 9).
The 5 vacant places can be filled in 5! ways.
Hence, required number of 6-digit numbers = 5! = 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.
2 different vowels and 2 different consonants are to be selected from the English alphabet.
Since there are 5 vowels in the English alphabet, number of ways of selecting 2 different vowels from the alphabet
= \({ }^{5} C_{2}=\frac{5 !}{2 ! 3 !}\) = 10
Since there are 21 consonants in the English alphabet, number of ways of selecting 2 different consonants from the alphabet
= \({ }^{21} \mathrm{C}_{2}=\frac{21 !}{2 ! 19 !}\)
= 210
Therefore, number of combinations of 2 different vowels and 2 different consonants = 10 × 210 = 2100.
Each of these 2100 combinations has 4 letters, which can be arranged among themselves in 4! ways.
Therefore, required number of words = 2100 × 4! = 50400.
Question 7.
In an examination, a question paper consists of 12 questions divided into two parts i.e., Part I and Part H, 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.
Student may select 8 questions according to following scheme
= 10 × 7 + 5 × 35 + 5 × 35
= 70 + 175 +175 = 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.
From a deck of 52 cards, 5-card combinations have to be made in such a way that in each selection of 5 cards, there is exactly one king.
In a deck of 52 cards, there are 4 kings.
1 king can be selected out of 4 kings in \({ }^{4} \mathrm{C}_{1}\) ways.
4 cards out of the remaining 48 cards can be selected in \({ }^{48} \mathrm{C}_{4}\) ways.
Thus, the required number of 5-card combinations is \({ }^{4} \mathrm{C}_{1} \times{ }^{48} \mathrm{C}_{4}\).
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 are to be seated in a row such that the women occupy the even places.
The 5 men can be seated in 5! ways.
For each arrangement, the 4 women can be seated only at the cross marked places (so that women occupy the even places).
M × M × M × M × M
Therefore, the women can be seated in 4! ways.
Thus, possible number of arrangements = 4! × 5!
= 24 × 120 = 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.
There are two cases :
(a) If the 3 students join the excursion party then the number of combinations will be P1 = C (22, 7).
(b) If the 3 students do not join the excursion party. Then the number of combinations P2 = C (22, 10).
If P is the combination of choosing the excursion party, then
P = P1 + P2
= C(22, 7) + C(22,10)
= \(\frac{22 !}{7 ! 15 !}+\frac{22 !}{10 ! 12 !}\)
= \(\frac{22 \times 21 \times 20 \times 19 \times 18 \times 17 \times 16 \times 15 !}{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 15 !}\) + \(\frac{22 \times 21 \times 20 \times 19 \times 18 \times 17 \times 16 \times 15 \times 14 \times 13 \times 12 !}{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 12 !}\)
= 170544 + 646646 = 817190.
Question 11.
In how many ways can the letters of the word ASSASSINATION be arranged so that all the S’s are together?
Answer.
In the given word ASSASSINATION, the letter A appears 3 times, S appears 4 times, I appears 2 times, N appears 2 times, and all the other letters appear only once.
Since all the words have to be arranged in such a way that all the Ss are together, SSSS is treated as a single object for the time being.
This single object together with the remaining 9 objects will account for 10 objects.
These 10 objects in which there are 3 As, 2 Is, and 2 Ns can be arranged in \(\frac{10 !}{3 ! 2 ! 2 !}\) ways.
Thus, required number of ways of arranging the letters of the given word = \(\frac{10 !}{3 ! 2 ! 2 !}\)
= \(\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 !}{3 ! \times 2 \times 2}\)
= 151200.