Class 11 Maths Ch-6 Permutations and Combinations

NCERT Solutions of Class 11 Maths Ch-6 Permutations and Combinations

Exercise 6.1

Question 1:
How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that
(i) repetition of the digits is allowed?
(ii) repetition of the digits is not allowed?
Answer
(i) The number of ways in which three-digit numbers can be formed from the given digits = 5×5×5 = 125
(ii) The number of ways in which three-digit numbers can be formed without repeating the given digits = 5×4×3= 60

Thank you for reading this post, don't forget to subscribe!

Question 2:
How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?
Answer
The required number of three digit even numbers is 3 × 6 × 6 = 108

Question 3:
How many 4-letter code can be formed using the first 10 letters of the English alphabet, if no letter can be repeated?
Answer
The required number of 4-letter code = 10 × 9 × 8 × 7 = 5040

Question 4:
How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?
Answer
The required number of 5-digit telephone numbers = 8 × 7 × 6 = 336

Question 5:
A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?
Answer
The required number of possible outcomes = 2 × 2 × 2 = 8

Question 6:
Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?
Answer
The required number of different signals = 5 × 4 = 20

---

NCERT Solutions of Class 11 Maths Ch-6 Permutations and Combinations

---

Exercise 6.2

Question 1:
Evaluate
(i) 8 ! (ii) 4 ! – 3 !
Answer
(i) 8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 40320
(ii) 4!-3! = (4 × 3!) – 3! = 3! (4-1) = 3 × 2 × 1 × 3 = 18

Question 2:
Is 3 ! + 4 ! = 7 ! ?
Answer
LHS
= 3! + 4! = (3 × 2 × 1) + (4 × 3 × 2 × 1)
= 6 + 24 = 30
RHS
7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040
LHS ≠ RHS
So, 3! + 4! ≠ 7!

Question 3:
Compute

Answer

Question 4:
If

find x.
Answer

Question 5:
Evaluate

when (i) n = 6, r = 2 (ii) n = 9, r = 5.
Answer

---

---

Exercise 6.3

Question 1:
How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?
Answer
The required number of 3-digit numbers = ⁹P₃.

Question 2:
How many 4-digit numbers are there with no digit repeated?
Answer
The thousands place of the 4-digit number is to be filled with any of the digits from 1 to 9 as the digit 0 cannot be included. Therefore, the number of ways in which thousands place can be filled is 9.
The hundreds, tens, and units place can be filled by any of the digits from 0 to 9.
Number of such 3-digit numbers = ⁹P₃ = 9!/(9 – 3)! = 9!/6!
= (9×8×7×6!)/6! = 9× 8× 7 = 504 
Thus, the required number of 4 – digit numbers = 504 × 9 = 4536

Question 3:
How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?
Answer
3-digit even numbers are to be formed using the digits 1, 2, 3, 4, 6, 7.
Then, units digits can be filled in 3 ways by any of the digits, 2, 4, or 6.
Number of ways of filling hundreds and tens place = ⁵P₂ = 5!/(5 – 2)! = 5!/3! = (5×4×3!)/3! = 20 
Thus, the required number of 3 – digit numbers = 3 × 20 = 60

Question 4:
Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of these will be even?
Answer
4-digit numbers are to be formed using the digits, 1, 2, 3, 4, and 5.
Therefore, required number of 4 digit numbers = ⁵P₄ = 5!/(5 – 4)! = 5!/1! = 1 × 2× 3× 4× 5 = 120.

Among the 4-digit numbers formed by using the digits, 1, 2, 3, 4, 5, even numbers end with either 2 or 4.
The number of ways in which units place is filled with digits is 2.
Number of ways of filling the remaining places ⁴P₃ = 4!/(4 – 3)! = 4!/1! = 4× 3×2×1 = 24 
Thus, by multiplication principle, the required number of even numbers = 24 × 2 = 48

Question 5:
From a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman assuming one person can not hold more than one position?
Answer
From a committee of 8 persons, a chairman and a vice chairman are to be chosen in such a way that one person cannot hold more than one position.
Thus, required number of ways = ⁸P₂ = 8!/(8 – 2)! = 8!/6!
= (8×7×6!)/6! = 8×7 = 56

Question 6:
Find n if ^n-1P₃ : ⁿP₄ = 1 : 9.
Answer

Question 7:
Find r if
(i) ⁵P_r = 2 ⁶P_r-1 (ii) ⁵P_r = ⁶P_r-1
Answer

⇒ (7 – r)(6 – r) = 12 
⇒ 42 – 6r – 7r + r² = 12
⇒ r² – 13r + 30 = 0 
⇒ r² – 3r – 10r + 30 = 0 
⇒ r(r – 3)-10(r – 3) = 0 
⇒ (r – 3)(r – 10) = 0
⇒ (r – 3) = 0 or (r – 10) = 0 
⇒ r = 3 or r = 10 
It is known that, ⁿP_r = n!/(n – r)!, where 0 ≤ r ≤ n 
0 ≤ r ≤ 5 
Hence, r ≠ 10
r = 3 

⇒ (7 – r)(6 – r) = 6 
⇒ 42 – 7r – 6r + r² – 6 = 0
⇒ r² – 13r + 36 = 0 
⇒ r² – 4r – 9r + 36 = 0 
⇒ r(r – 4)-9(r – 4) = 0 
⇒ (r – 4)(r – 9) = 0 
⇒ (r – 4) = 0 or (r – 9) = 0 
⇒ r = 4 or r = 9
It is known that, ⁿP_r = n!/(n – r)!, where 0 ≤ r ≤ n 0 ≤ r ≤ 5 
Hence, r ≠ 9
r = 4

Question 8:
How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once?
Answer
The required number of words that can be formed = ⁸P₈ = 8! = 40320

Question 9:
How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if.
(i) 4 letters are used at a time, (ii) all letters are used at a time, (iii) all letters are used but first letter is a vowel?
Answer
There are 6 different letters in the word MONDAY.
(i) The required number of words that can be formed using 4 letters at a time is 

(ii) The required number of words that can be formed when all letters are used at a time = ⁶P₆ = 6!
= 6 × 5 × 4 × 3 × 2 ×1 = 720.

(iii) Vowels = O, A
There are 2 different vowels, which have to occupy the rightmost place of the words formed.
This can be done only in 2 ways.
The remaining five places are to be filled by the remaining 5 letters. This can be done in 5! ways.
Thus, the required number of words that can be formed = 5! × 2 = 120 × 2 = 240

uestion 10:
In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s not come together?
Answer
In the given word MISSISSIPPI,
I appears 4 times,
S appears 4 times, and
P appears 2 times.
Therefore, number of distinct permutations of the letters in the given word

There are 4 Is in the given word. When they occur together, they are treated as a single object  for the time being. This single object together with the remaining 7 objects will account for 8 objects.
These 8 objects in which there are 4 Ss and 2 Ps can be arranged in 8!/4!2! ways i.e., 840 ways. 
Number of arrangements where all is occur together = 840 
Thus, number of distinct permutations of the letters in MISSISSIPPI in which four is do not come together
= 34650 – 840 = 33810

Question 11:
In how many ways can the letters of the word PERMUTATIONS be arranged if the
(i) words start with P and end with S, (ii) vowels are all together, (iii) there are always 4 letters between P and S?
Answer
In the word PERMUTATIONS, there are 2 Ts and all the other letters appear only once.
(i) If P and S are fixed at the extreme ends (P at the left end and S at the right end), then 10 letters are left.
Hence the required number of arrangements = 10!/2! = 1814400

(ii) There are 5 vowels in the given word, each appearing only once.
Therefore, the required number of arrangements in this case = (8!/2!) × 5!= 2419200.

(iii) The letters have to be arranged in such a way that there are always 4 letters between P and S.
The places of P and S are fixed. The remaining 10 letters in which there are 2 Ts can be arranged in 10!/2! ways .
Also, the letters P and S can be placed such that there are 4 letters between them in 2×7 = 14 ways
Therefore, by multiplication principle, required number of arrangements in this case = (10!/2!)×14 = 25401600

---

NCERT Solutions of Class 11 Maths Ch-6 Permutations and Combinations

---

Exercise 6.4

Question 1:
If ⁿC₈ = ⁿC₂, find ⁿC₂.
Answer
ⁿC_a = ⁿC_b ⇒ a = b or n = a+ b
Therefore,
ⁿC₈ = ⁿC₂ ⇒ n = 8 + 2 = 10 
∴ ⁿC₂ = ¹⁰C₂ = 

Question 2:
Determine n if
(i) ²ⁿC₃ : ⁿC₃ = 12 : 1 (ii) ²ⁿC₃ : ⁿC₃ = 11 : 1
Answer
(i) 

⇒ 2n – 1 = 3(n – 2) 
⇒ 2n – 1 = 3n – 6 
⇒ 3n – 2n = -1 + 6 
⇒ n = 5

(ii)

⇒ 4(2n – 1) = 11(n – 2)
⇒ 8n – 4 = 11n – 22 
⇒ 11n – 8n = -4 + 22 
⇒ 3n = 18 
⇒ n = 6

Question 3:
How many chords can be drawn through 21 points on a circle?
Answer
For drawing one chord on a circle, only 2 points are required.
Thus, the required number of chords = ²¹C₂ = 21!/2!(21 -2)! = 21!/2!19! = (21×20)/2 = 210

Question 4:
In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?
Answer
A team of 3 boys and 3 girls is to be selected from 5 boys and 4 girls. 
3 boys can be selected from 5 boys in  ⁵C₃  ways .
3 girls can be selected from 4 girls in ⁴C₃  ways. 
Therefore, the number of ways in which a team of 3 boys and  3 girls can be selected = ⁵C₃ × ⁴C₃  
= (5!/3!2!) × (4!/3!1!) 

Question 5:
Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour.
Answer
There are a total of 6 red balls, 5 white balls, and 5 blue balls.
9 balls have to be selected in such a way that each selection consists of 3 balls of each colour.
Here,
3 balls can be selected from 6 red balls in ⁶C₃ ways.
3 balls can be selected from 5 white balls in ⁵C₃ ways.
3 balls can be selected from 5 blue balls in ⁵C₃ ways.
Thus, the required number of ways of selecting 9 balls

Question 6:
Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination.
Answer
In a deck of 52 cards, there are 4 aces.
Then, one ace can be selected in ⁴C₁ ways and the remaining 4 cards can be selected out of the 48 cards in ⁴⁸C₄ ways.
Thus, the required number of 5 card combinations 

Question 7:
In how many ways can one select a cricket team of eleven from 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers?
Answer
Out of 17 players, 5 players are bowlers.
A cricket team of 11 players is to be selected in such a way that there are exactly 4 bowlers.
4 bowlers can be selected in ⁵C₄ ways and the remaining 7 players can be selected out of the 12 players in ¹²C₇ ways.
Thus, the required number of ways of selecting cricket team 

Question 8:
A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.
Answer
There are 5 black and 6 red balls in the bag.
2 black balls can be selected out of 5 black balls in ⁵C₂ ways and 3 red balls can be selected out of 6 red balls in ⁶C₃ ways.
Thus, the required number of ways of selecting 2 black and 3 red balls 

Question 9:
In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?
Answer
There are 9 courses available out of which, 2 specific courses are compulsory for every student.
Therefore, every student has to choose 3 courses out of the remaining 7 courses. This can be chosen in ⁷C₃ ways.
Thus, required number of ways of choosing the programme

---

NCERT Solutions of Class 11 Maths Ch-6 Permutations and Combinations

---

Miscellaneous Exercise on Chapter 6

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! = 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 ²C₂ = 2! 
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, the required number of words = 2! × 5! × 3! = 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, the required number of ways = ⁴C₃ × ⁹C₄ = (4!/3!1!) × (9!/4!5!) 


(ii) Since at least 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 ⁴C₃ × ⁹C₄ ways. 
4 girls and 3 boys can be selected in ⁴C₃ × ⁹C₄ ways.  
Therefore, the required number of ways = ⁴C₃ × ⁹C₄ + ⁴C₃ × ⁹C₄  = 504 + 84 = 588 

(iii) Since at most 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 ways. 
1 girl and 6 boys can be selected in ⁴C₁ × ⁹C₆ ways . 
No girl and 7 boys can be selected in ⁴C₀ × ⁹C₇ ways . 
Therefore, in this case, required number of ways . 

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 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 = ⁵C₂ = 5!/2!3! = 10 
Since there are 21 consonants in the English alphabet, number of ways of selecting 2 different consonants from the alphabet = ²¹C₂ = 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 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
It is given that the question paper consists of 12 questions divided into two parts – Part I and Part II, containing 5 and 7 questions, respectively.
A student has to attempt 8 questions, selecting 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 questions from part I and 5 questions from part II can be selected in ⁵C₃ × ⁷C₅ 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 × ⁷C₃ ways.
Thus, required number of ways of selecting questions

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 is 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
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
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, the 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.
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! = 151200

---

NCERT Solutions of Class 11 Maths Ch-6 Permutations and Combinations

---

BrightWay Coaching Academy