Class 11 - Mathematics
Permutations Combinations - Exercise 7-Misc
Top Block 1
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:
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! = 30 * 120 = 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:
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 five 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!
= 2 * 120 * 6
= 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) at most 3 girls?
Answer:
Given number of boys = 9
Number of girls = 4
Now, A committee of 7 has to be formed from 9 boys and 4 girls.
- If in committee consist of exactly 3 girls:
4C3 * 9C4
= {4! / (3! * 1!)} * {9! / (4! * 5!)}
= {(4*3!) /3!} * {(9*8*7*6*5!) / (4! * 5!)}
= 4 * {(9*8*7*6*) / 4!}
= {4 * (9*8*7*6*)} / (4*3*2*1)
= 9*8*7
= 504
- If committee consists of at least 3 girls:
4C3 * 9C4 + 4C4 * 9C3
= [{4! / (3! * 1!)} * {9! / (4! * 5!)}] + 9C3
= [{(4*3!) /3!} * {(9*8*7*6*5!) / (4! * 5!)}] + 9! /(3! * 6!)
= [4 * {(9*8*7*6*) / 4!}] + (9*8*7*6!)/(3! * 6!)
= [{4 * (9*8*7*6*)} / (4*3*2*1)] + (9*8*7)/3!
= (9*8*7) + (9*8*7)/(3*2*1)
= 504 + (504/6)
= 504 + 84
= 588
- If committee consists of at most 3 girls:
= 4C0 * 9C7 + 4C1 * 9C6 + 4C2 * 9C5 + 4C3 * 9C4
= 36 + 336 + 756 + 504
= 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:
In the given word EXAMINATION, there are 11 letters out of which, A, I, and N appear 2 times
and all the other letters appear only once.
The words that will be listed before the words starting with E in a dictionary will be the words
that start with A only.
Therefore, to get the number of words starting with A, the letter A is fixed at the extreme left
position, and then the remaining 10 letters taken all at a time are rearranged.
Since there are 2 Is and 2 Ns in the remaining 10 letters,
Number of words starting with A =10!/(2! * 2!) = 907200
Thus, the required numbers of words is 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 the unit digit of the number is 0.
Given digits are: 0, 1, 3, 5, 7, 9
Now we fix digit 0 at unit place of the number.
Remaining 5 digits can be arranged in 5! ways
So, total 6-digit numbers which are divisible by 10 = 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:
Number of vowels = 5
Number of consonants = 21
Now we have to form words with 2 different vowels and 2 different consonants.
Number of words = 5C2 * 21C2 * 4! (4! is because we can arrange the letters in 4! ways)
= (5*4)/2 * (21*20)/2 * 24 (4! = 24)
= 20/2 * 420/2 * 24
= 10*210*24
= 50400
So, total number of words = 50400
Mddle block 1
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, total number of questions = 12
Part 1 contains 5 questions and part 2 contain 7 questions.
A student has to select 8 questions all in which there is at least 3 questions from each part.
Now different possibilities can occur.
Part1 (5 questions) Part2 (7 questions)
3 5
4 4
5 3
Now total number of ways is
(5C3 * 7C5) + (5C4 * 7C4) + (5C5 * 7C3)
= (5C2 * 7C2) + (5C1 * 7C3) + (5C0 * 7C3)
= {(5 * 4 * 7 * 6)/(2 * 2)} + {(5 * 7 * 6 * 5)/(3 * 2)} + {(7 * 6 * 5)/(3 * 2)}
= 210 + 175 + 35
= 420
So, the total number of ways a student can select the questions is 420
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 4C1 ways.
4 cards out of the remaining 48 cards can be selected in 48C4 ways.
Thus, the required number of 5-card combinations = 4C1 * 48C4
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:
4 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 :
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 the three students join.
Then, the remaining 7 students can be chosen from the remaining 22 students in 22C7 ways.
Case II: None of the three students join.
Then, 10 students can be chosen from the remaining 22 students in 22C10 ways.
Thus, required number of ways of choosing the excursion party = 22C7 + 22C10
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 10!/(3! * 2! * 2!)
ways.
Thus, required number of ways of arranging the letters of the given word
= 10!/(3! * 2! * 2!)
= (10 * 9 * 8 * 7 * 6 * 5 * 4 * 3!/(3! * 2 * 2)
= 10 * 9 * 8 * 7 * 6 * 5
= 151200
Bottom Block 3
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