Referring to the AP Inter 1st Year Maths Study Material Chapter 6 Permutations and Combinations Exercise 6e Solutions makes it easier to understand complex problems.
AP Inter 1st Year Maths Permutations and Combinations Solutions Exercise 6e
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) atmost 3 girls.
Solution:
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
= 4C3 × 9C4
= \(\frac{4!}{3!1!} \times \frac{9!}{4!5!}\)
= 4 × \(\frac{9 \times 8 \times 7 \times 6 \times 5!}{4 \times 3 \times 2 \times 1 \times 5!}\)
= 508
ii) Since atleast 3 girls are to be there in every committee. The committee can consist of
a) 3 girls and 4 boys
b) 4 girls and 3 boys
3 girls and 4 boys can be selected in 4C3 × 9C4 ways.
4 girls and 3 boys can be selected in 4C2 × 9C5 ways.
Therefore, in this case, required number of ways.
= 4C3 × 9C4 + 4C4 × 9C3
= 504 + 84
= 588
Therefore, in this case, required number of ways.
= 4C3 × 9C4 + 4C4 × 9C3
= 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 4C3 × 9C4 ways.
2 girls and 5 boys can be selected in 4C2 × 9C5
1 girl and 6 boys can be selected in 4C1 × 9C6
No girl and 7 boys can be selected in 4C0 × 9C7 ways.
Therefore, in this case, required number of ways.
= 4C3 × 9C4 + 4C2 × 9C5 + 4C1 × 9C6 + 4C0 × 9C7
= \(\frac{4!}{3!1!} \times \frac{9!}{4!5!}+\frac{4!}{2!2!} \times \frac{9!}{5!4!}+\frac{4!}{1!3!} \times \frac{9!}{6!3!}+\frac{4!}{0!4!} \times \frac{9!}{7!2!}\)
= 504 + 756 + 336 + 36
= 1632
Question 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:
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 = M × M × M × M × M.
Therefore, the women can be seated in 4! ways.
Hence, possible number of arrangements = 4! × 5! = 24 × 120 = 2880.
Question 3.
In how many ways can be letters of the word ASSASSINATION be arranged so that all the S’s are together ?
Solution:
In the given word ASSASSINATION, the letter A appear 3 times.
S appears 4 times, I appear 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 ‘S’ are together, SSSS is treated as a single objects for the time being. This single object together with the remaining 9 objects will account for 10 objects.
The number of arrangements of 10 objects in which there are 3’A’, 2T and 2’N’.
\(\frac{10!}{3!2!2!}=\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3!}{3!\times 2 \times 1 \times 2 \times 1}\) = 151200.
Hence, the required number of ways of arranging the letters of the given word is 151200.
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:
In the word DAUGHTER, there are 3 vowels namely A, U and E and 5 consonants, namely D, G, H, T and R.
Number of ways of selecting 2 vowels out of 3 vowels = 3C2 = 3.
Number of ways of selecting 3 consonants out of 5 consonants = 5C3 =10. Therefore, number of combinations of 2 vowels and 3 consonants = 3 × 10 = 30.
Each of these 30 combinations of 2 vowels and 3 consonants can be arranged among themselves in 5! ways.
Hence, required number of different words = 30 × 5! = 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 ?
Solution:
In the word EQUATION, there are 5 vowels, namely, A, E, I, O and U and 3 consonants namely Q, T and N.
Since all the vowels and consonants have to occur together, both (AEIOU) and (QTN) can be assumed as single objects. Then, the permutations of these 2 objects taken all at a time are counted.
This number would be 2P2 = 2!
Corresponding to each of these permutations, there are 5! Permutations of the 5 vowels taken all at a time and 3! Permutations of the 3 consonants taken all at a time.
Hence, by multiplication principle, required number of words = 2! × 5! × 3! = 1440.
Question 3.
If the different permutations of all the letters of the word EXAMINATION are listed in a dictionary, how many words are there in this list before first word start with E ?
Solution:
In the given word EXAMINATION there are 11 letters out of which A, I and N appear 2 times and all the other letters appears only once.
The words that will be listed before the word starting with E in a dictionary will be word that start with A. Therefore to get the number of words starting with E in a dictionary will be the words that start with A since there are 2T and 2 ’N’ in the remaining 10 letters.
Number of words starting with A = \(\frac{10!}{2!2!}=\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2!}{2 \times 1 \times 2!}\)
= 907200
Hence, the required number of words are 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:
A number is divisible by 10 if its unit’s digits is 0.
Therefore, 0 is fixed at the unit’s 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-digits = 5! = 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:
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 = 5C2 = \(\frac{5!}{2!3!}=\frac{5 \times 4 \times 3!}{2 \times 1 \times 3!}\) = 10
Since there are 21 consonants in the English alphabet.
Number of ways of selecting 2 different consonants from the alphabet
= 21C2 = \(\frac{21!}{2!9!}=\frac{21 \times 20 \times 19!}{2 \times 1 \times 19!}\) = 210
Therefore, number of combinations of 2 different vowels and 2 different consonants = 10 x 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.
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:
It is given that the question paper consists of 12 questions divided into two parts – part I and and part II containing 5 and 7 questions, respectively.
A student has to attempts 8 questions at least 3 from each part. This can be done as follows.
a) 3 questions from part I and 5 questions from part II
b) 4 questions from part I and 4 questions from part II
c) 5 questions from part I and 3 questions from part II
3 questios from part I and 5 questions from part II can be selected in 5C3 × 7C4 ways.
4 questions from part I and 4 questions from part II can be selected in 5C4 × 7C4 ways.
5 questions from part I and 3 questions from part II can be selected in 5C5 × 7C2 ways
= 5C3 × 7C4 + 5C4 × 7C4 + 5C5 × 7C2
Question 3.
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 ?
Solution:
From the class of 25 students, 10 are to be chosen for an excursion party.
Since there are 3 students who decide that either all of them will join or none of them will join, there are two cases.
Case-I : All of three students join.
Then, 10 students can be chosen from the remaining 22 students 22C7 ways.
Case-II : None of the three students join.
Then, 10 students can be chosen from the remaining 22 students in 22C10 ways. Hence, the required number of ways of choosing the excursion party is 22C7 + 22C10