Students can go through AP Inter 1st Year Maths Notes 6th Lesson Permutations and Combinations will help students in revising the entire concepts quickly.
Permutations and Combinations Class 11 Notes AP Inter 1st Year Maths 6th Lesson
→ Fundamental principle of counting. If an event can occur in m different ways, following which another event can occur in n different ways, then the total number of occurrences of the events in the given order is m × n.
→ The number of permutations of n different things taken r at a time, where repetition is not allowed, is denoted by nPr and is given by \({ }^n P_r=\frac{n!}{(n-r)!}\), where 0 ≤ r ≤ n.
→ n! = 1 × 2 × 3 × … × n
→ n! = n × (n – 1)!
→ The number of permutations of n different things, taken r at a time, where repetition is allowed, is nr.
→ The number of permutations of n objects taken all at a time, where p1 objects are of first kind, p2 objects are of the second kind,…, pk objects are of the kth kind and rest, if any, are all different is \(\frac{n!}{p_{1}!p_{2}!\ldots . p_{k}!}\).
→ The number of combinations of n different things taken r at a time, denoted by nCr, is given by nCr, is given by \({ }^n C_r=\frac{n!}{r!(n-r)!}\), 0 ≤ r ≤ n.
→ nCr = nCs then n = r + s or r = s for 0 ≤ r, s < n.
→ If n, r are integers with 0 ≤ r ≤ n, then nCr = nC(n-r)
→ If 1 ≤ r ≤ n, then nCr-1 + nCr = (n+1)Cr
Example Problems
Question 1.
Find the number of 4-letter words, with or without meaning, which can be formed out of the letters of the word ROSE, where the repetition of the letters is not allowed.
Solution:
There are as many words as there are ways of filling in 4 vacant places 
with the 4 letters, keeping in mind that repetition is not allowed.
The first place can be filled in 4 different ways by any one of the 4 letters R, O, S, E.
Following which, the second place can be filled in by any one of the remaining 3 letters in 3 different ways, following which the third place can be filled in 2 different ways; following which, the fourth place can be filled in 1 way.
Thus, the number of ways in which the 4 places can be filled, by the multiplication principle, is 4 × 3 × 2 × 1 = 24.
Hence, the required number of words is 24.
Note: If the repetition of the letters was allowed, how many words can be formed?
One can easily understand that each of the 4 vacant places can be filled in succession in 4 different ways.
Hence, the required number of words = 4 × 4 × 4 × 4 = 256.
Question 2.
Given 4 flags of different colours, how many different signals can be generated, if a signal requires the use of 2 flags, one below the other?
Solution:
There will be as many signals as there are ways of filling in 2 vacant places 
in succession by the 4 flags of different colours.
The upper vacant place can be filled in 4 different ways by any one of the 4 flags; following which, the lower vacant place can be filled in 4 different ways by any one of the 4 flags: following which, the lower vacant place can be filled in 3 different ways by any one of the remaining 3 different flags.
Hence, by the multiplication principle, the required number of signals = 4 × 3 = 12.
Question 3.
How many 2-digit even numbers can be formed from the digits 1, 2, 3, 4, 5 if the digits can be repeated?
Solution:
There will be as many 2-digit numbers as there are ways of filling 2 vacant places 
in succession by the five given digits. Here, in this case, we start filling in the unit’s place, because the options for this place are 2 and 4 only, and this can be done in 2 ways; following which the ten’s place can be filled by any of the 5 digits in 5 different ways, as the digits can be repeated. Therefore, by the multiplication principle, the required number of two-digit even numbers is 2 × 5 = 10.
Question 4.
Find the number of different signals that can be generated by arranging at least 2 flags in order (one below the other) on a vertical staff, if five different flags are available.
Solution:
A signal can consist of either 2 flags, 3 flags, 4 flags, or 5 flags.
Now, let us count the possible number of signals consisting of 2 flags, 3 flags, 4 flags, and 5 flags separately and then add the respective numbers.
There will be as many 2-flag signals as there are ways of filling in 2 vacant places 
in succession by the 5 flags available.
By the Multiplication rule, the number of ways is 5 × 4 = 20.
Similarly, there will be as many 3-flag signals as there are ways of filling in 3 vacant places 
in succession by the 5 flags.
The number of ways is 5 × 4 × 3 = 60 continuing the same way, we find that
The number of 4 flag signals = 5 × 4 × 3 × 2 = 120.
and the number of 5 flag signals = 5 × 4 × 3 × 2 × 1 = 120.
Therefore, the required no of signals = 20 + 60 + 120 + 120 = 320.
Question 5.
Evaluate
(i) 5!
(ii) 7!
(iii) 7! – 5!
Solution:
(i) 5! = 1 × 2 × 3 × 4 × 5 = 120
(ii) 7! = 1 × 2 × 3 × 4 × 5 × 6 × 7 = 5040
(iii) 7! – 5! = 5040 – 120 = 4920
Question 6.
Compute
(i) \(\frac{7!}{5!}\)
(ii) \(\frac{12!}{(10!)(2!)}\)
Solution:
(i) We have \(\frac{7!}{5!}=\frac{7 \times 6 \times 5!}{5!}\)
= 7 × 6
= 42
(ii) \(\frac{12!}{(10!)(2!)}=\frac{12 \times 11 \times(10!)}{(10!) \times(2)}\)
= 6 × 11
= 66
Question 7.
Evaluate \(\frac{n!}{r!(n-r)!}\), when n = 5, r = 2.
Solution:
We have to evaluate \(\frac{5!}{2!(5-2)!}\) (Since n = 5, r = 2)
We have \(\frac{5!}{2!(5-2)!}=\frac{5!}{2!\times 3!}=\frac{5 \times 4}{2}\) = 10
Question 8.
If \(\frac{1}{8!}+\frac{1}{9!}=\frac{x}{10!}\), find x.
Solution:
We have \(\frac{1}{8!}+\frac{1}{9 \times 8!}=\frac{x}{10 \times 9 \times 8!}\)
Therefore \(1+\frac{1}{9}=\frac{x}{10 \times 9}\)
⇒ \(\frac{10}{9}=\frac{x}{10 \times 9}\)
⇒ x = 100
Question 9.
Find the number of permutations of the letters of the word ALLAHABAD.
Solution:
Here, there are 9 objects (letters), of which there are 4 A’s, 2 L’s, and the rest are all different.
Therefore, the required number of arrangements = \(\frac{9!}{4!2!}\)
= \(\frac{5 \times 6 \times 7 \times 8 \times 9}{2}\)
= 7560
Question 10.
How many 4-digit numbers can be formed by using the digits 1 to 9 if repetition of digits is not allowed?
Solution:
Here, order matters. For example, 1234 and 1324 are two different numbers.
Therefore, there will be as many 4-digit numbers as there are permutations of 9 different digits taken 4 at a time.
Therefore, the required 4-digit numbers = 9P4
= \(\frac{9!}{(9-4)!}\)
= \(\frac{9!}{5!}\)
= 9 × 8 × 7 × 6
= 3024
Question 11.
How many numbers lying between 100 and 1000 can be formed with the digits 0, 1, 2, 3, 4, 5 if the repetition of the digits is not allowed?
Solution:
Every number between 100 and 1000 is a 3-digit number.
We first have to count the permutations of 6 digits taken 3 at a time.
This number would be 6P3.
But, these permutations will include those also where 0 is at the 100’s place.
For example, 092, 042, etc., are such numbers that are actually 2-digit numbers, and hence the number of such numbers has to be subtracted from 6P3 to get the required number.
To get the number of such members, we fix 0 at the 100’s place and rearrange the remaining 5 digits, taking 2 at a time. This number is 5P2.
So The required number = 6P3 – 5P2
= \(\frac{6!}{3!}-\frac{5!}{3!}\)
= 4 × 5 × 6 – 4 × 5
= 100
Question 12.
Find the value of n such that
(i) nP5 = 42 nP3, n > 4
(ii) \(\frac{{ }^n P_4}{{ }^{n-1} P_4}=\frac{5}{3}\), n > 4
Solution:
(i) Given that nP5 = 42 nP3
or n(n – 1) (n – 2) (n – 3) (n – 4) = 42 n(n – 1) (n – 2)
Since n > 4 so n(n – 1) (n – 2) ≠ 0
Therefore, by dividing both sides by n(n – 1) (n – 2), we get
(n – 3) (n – 4) = 42
⇒ n2 – 7n – 30 = 0
⇒ n2 – 10n + 3n – 30
⇒ (n – 10) (n + 3) = 0
⇒ n – 10 = 0 or n + 3 = 0
⇒ n = 10 or n = -3
As n cannot be negative, n = 10.
(ii) Given that \(\frac{{ }^n P_4}{{ }^{n-1} P_4}=\frac{5}{3}\)
Therefore, 3n (n – 1) (n – 2) (n – 3) = 5(n – 1) (n – 2) (n – 3) (n – 4)
⇒ 3n = 5(n – 4) [as (n – 1) (n – 2) (n – 3) ≠ 0, n > 4]
⇒ n = 10
Question 13.
Find r, if 5 4Pr = 6 5Pr-1.
Solution:
We have 5 4Pr = 6 5Pr-1
⇒ 5 × \(\frac{4!}{(4-r)!}\) = 6 × \(\frac{5!}{(5-r+1)!}\)
⇒ \(\frac{5!}{(4-r)!}=\frac{6 \times 5!}{(5-r+1)(5-r)(5-r-1)!}\)
⇒ (6-r) (5-r) = 6
⇒ r2 – 11r + 24 = 0
⇒ r2 – 8r – 3r + 24 = 0
⇒ (r – 8)(r – 3) = 0
⇒ r = 8 or r = 3
Hence r = 8, 3
Question 14.
Find the number of different 8-letter arrangements that can be made from the letters of the word DAUGHTER so that
(i) All vowels occur together.
(ii) All vowels do not occur together.
Solution:
(i) There are 8 different letters in the word DAUGHTER, in which there are 3 vowels, namely, A, U, and E.
Since the vowels have to occur together, we can, for the time being, assume them as a single object (AUE).
This single object, together with 5 remaining letters (objects), will be counted as 6 objects.
Then we count permutations of these 6 objects taken all at a time.
This number would be 6P6 = 6!
Corresponding to each of these permutations, we shall have 3! permutations of the three vowels, A, U, E, taken all at a time.
Hence, by the multiplication principle, the required number of permutations = 6! × 3! = 4320.
(ii) If we have to count those permutations in which all vowels are never together, we first have to find all possible arrangements of 8 letters taken all at a time, which can be done in 8! ways. Then, we have to subtract from this number the number of permutations in which the vowels are always together.
Therefore, the required number = 8! – 6! × 3!
= 6! (7 × 8 – 6)
= 2 × 6! (28 – 3)
= 50 × 6!
= 50 × 720
= 36000
Question 15.
In how many ways can 4 red, 3 yellow, and 2 green discs be arranged in a row if the discs of the same colour are indistinguishable?
Solution:
Total number of discs is 4 + 3 + 2 = 9.
Out of 9 discs, 4 are of the first kind (red), 3 are of the second kind (yellow), and 2 are of the third kind (green).
Therefore, the number of arrangements \(\frac{9!}{4!3!2!}\) = 1260.
Question 16.
Find the number of arrangements of the letters of the word INDEPENDENCE. In how many of these arrangements,
(i) Do the words start with P
(ii) Do all the vowels always occur together
(iii) Do the vowels never occur together
(iv) Do the words begin with I and end in P?
Solution:
There are 12 letters, of which N appears 3 times, E appears 4 times, and D appears 2 times, and the rest are all different.
Therefore The required number of arrangements = \(\frac{12!}{3!4!2!}\) = 1663200
(i) Let us fix P at the extreme left position and then count the arrangements of the remaining 11 letters.
Therefore, the required number of words starting with P = \(\frac{11!}{3!2!4!}\) = 138600
(ii) There are 5 vowels in the given word, which are 4 Es and 1I.
Since they have to always occur together, we treat them as a single object EEEEI for the time being.
This single object, together with 7 remaining objects, will account for 8 objects.
These 8 objects, in which there are 3 Ns and 2 Ds, can be rearranged in \(\frac{8!}{3!2!}\) ways.
Corresponding to each of these arrangements, the 5 vowels E, E, E, E, and I can be rearranged in \(\frac{5!}{4!}\) ways.
Therefore, by multiplication principle, the required number of arrangements = \(\frac{8!}{3!2!} \times \frac{5!}{4!}\) = 16800
(iii) The required number of arrangements = the total number of arrangements (without any restriction) – the number of arrangements where all the vowels occur together
= 1663200 – 16800
= 1646400
(iv) Let us fix I and P at the extreme ends (I at the left end and P at the right end).
We are left with 10 letters.
Hence, the required number of arrangements = \(\frac{10!}{3!2!4!}\) = 12600
Question 17.
If nC9 = nC8, find nC17.
Solution:
We have nC9 = nC8
⇒ \(\frac{n!}{9!(n-9)!}=\frac{n!}{(n-8)!8!}\)
⇒ \(\frac{1}{9}=\frac{1}{n-8}\)
⇒ n – 8 = 9
⇒ n = 17
Therefore, nC17 = 17C17 = 1
Question 18.
A committee of 3 persons is to be constituted from a group of 2 men and 3 women. In how many ways can this be done? How many of these committees would consist of 1 man and 2 women?
Solution:
Here, order does not matter. Therefore, we need to count combinations.
There will be as many committees as there are combinations of 5 different persons taken 3 at a time.
Hence, the required number of ways = 5C3
= \(\frac{5!}{3!2!}\)
= \(\frac{4 \times 5}{2}\)
= 10
Now, 1 man can be selected from 2 men in 2C1 ways, and 2 women can be selected from 3 women in 3C2 ways.
Therefore, the required number of committees = 2C1 × 3C2
= \(\frac{2!}{1!1!} \times \frac{3!}{2!1!}\)
= 6
Question 19.
What is the number of ways of choosing 4 cards from a pack of 52 playing cards? In how many of these
(i) four cards are of the same suit,
(ii) foureards belong to four different suits,
(iii) are face cards,
(iv) two are red cards, and two are black cards,
(v) cards of the same colour?
Solution:
There will be as many ways of choosing 4 cards from 52 cards as there are combinations of 52 different things, taken 4 at a time.
Therefore, the required number of ways = 52C4
= \(\frac{52!}{4!48!}\)
= \(\frac{49 \times 50 \times 51 \times 52}{2 \times 3 \times 4}\)
= 270725.
(i) There are four suits: diamond, club, spade, heart, and there are 13 cards of each suit.
Therefore, there are 13C4 ways of choosing 4 diamonds.
Similarly, there are 13C4 ways of choosing 4 clubs, 13C4 ways of choosing 4 spades, and 13C4 ways of choosing 4 hearts.
Therefore The required number of ways = 13C4 + 13C4 + 13C4 + 13C4
= 4 × \(\frac{13!}{4!9!}\)
= 2860
(ii) There are 13 cards in each suit.
Therefore, there are 13C1 ways of choosing 1 card from 13 cards of diamonds, 13C1 ways of choosing 1 card from 13 cards of hearts, 13C1 ways of choosing 1 card from 13 cards of clubs, 13C1 ways of choosing 1 card from 13 cards of spades.
Hence, by multiplication principle, the required number of ways = 13C1 × 13C1 × 13C1 × 13C1 = 134
(iii) There are 12 face cards, and 4 are to be selected out of these 12 cards.
This can be done in 12C4 ways.
Therefore, the required number of ways = \(\frac{12!}{4!8!}\) = 495
(iv) There are 26 red cards and 26 black cards.
Therefore, the required number of ways = 26C2 × 26C2
= \(\left(\frac{26!}{2!24!}\right)^2\)
= (325)2
= 105625
(v) 4 red cards can be selected out of 26 red cards in 26C4 ways.
4 black cards can be selected out of 26 black cards in 26C4 ways.
Therefore, the required number of ways = 26C4 + 26C4
= 2 × \(\frac{26!}{4!22!}\)
= 29900
Question 20.
How many words, with or without meaning, each of 3 vowels and 2 consonants can be formed from the letters of the word INVOLUTE?
Solution:
In the word INVOLUTE, there are 4 vowels, namely, I, O, E, and U, and 4 consonants, namely N, V, L, and T.
The number of ways of selecting 3 vowels out of 4 = 4C3 = 4
The number of ways of selecting 2 consonants out of 4 = 4C4 = 6
Therefore, the number of combinations of 3 vowels and 2 consonants is 4 × 6 = 24.
Now, each of these 24 combinations has 5 letters, which can be arranged among themselves in 5! ways.
Therefore, the required number of different words is 24 × 5! = 2880.
Question 21.
A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has
(i) No girl?
(ii) At least one boy and one girl?
(iii) At least 3 girls?
Solution:
(i) Since the team will not include any girls, therefore, only boys are to be selected.
5 boys out of 7 boys can be selected in 7C5 ways.
Therefore, the required number of ways = 7C5
= \(\frac{7!}{5!2!}\)
= \(\frac{6 \times 7}{2}\)
= 21
(ii) Since at least one boy and one girl are to be there in every team.
Therefore, the team can consist of
(a) 1 boy and 4 girls
(b) 2 boys and 3 girls
(c) 3 boys and 2 girls
(d) 4 boys and 1 girl
1 boy and 4 girls can be selected in 7C1 × 4C4 ways.
2 boys and 3 girls can be selected in 7C2 × 4C3 ways.
3 boys and 2 girls can be selected in 7C3 × 4C2 ways.
4 boys and 1 girl can be selected in 7C4 × 4C1 ways.
Therefore, the required number of ways = 7C1 × 4C4 + 7C2 × 4C3 + 7C3 × 4C2 + 7C4 × 4C1
= 7 + 84 + 210 + 140
= 441
(iii) Since the team has to consist of atleast 3 girls, the team can consist of (a) 3 girls and 2 boys, or (b) 4 girls and 1 boy.
Note that the team cannot have all 5 girls, because the group has only 4 girls.
3 girls and 2 boys can be selected in 4C3 × 7C2 ways.
4 girls and 1 boy can be selected in 4C4 × 7C1 ways.
Therefore, the required number of ways = 4C3 × 7C2 + 4C4 × 7C1
= 84 + 7
= 91
Question 22.
Find the number of words with or without meaning which can he made using all the letters of the word AGAIN. If these words are written as in a dictionary, what will be the 50th word?
Solution:
There are 5 letters in the word AGAIN, in which A appears 2 times.
Therefore, the required number of words = \(\frac{5!}{2!}\) = 60.
To get the number of words starting with A, we fix the letter A at the extreme left position, we then rearrange the remaining 4 letters taken all at a time.
There will be as many arrangements of these 4 letters taken 4 at a time as there are permutations of 4 different things taken 4 at a time.
Hence, the number of words starting with A = 4! = 24.
Then starting with G, the number of words = \(\frac{4!}{2!}\) = 12, as after placing G at the extreme left position, we are left with the letters A, A, I, and N.
Similarly, 12 words start with the next letter I.
Total number of words so far obtained = 24 + 12 + 12 = 48.
The 49th word is NAAGI.
The 50th word is NAAIG.
Question 23.
How many numbers greater than 1000000 can be formed by using the digits 1, 2, 0, 2, 4, 2, 4?
Solution:
Since 1000000 is a 7-digit number, and the number of digits to be used is also 7.
Therefore, the numbers to be continued will be 7-digit only.
Also, the numbers have to be greater than 1000000, so they can begin with either 1, 2, or 4.
The number of numbers beginning with 1 = \(\frac{6!}{3!2!}=\frac{4 \times 5 \times 6}{2}\) = 60,
as when 1 is fixed at the extreme left position, the remaining digits to be rearranged will be 0, 2, 2, 2, 4, 4 in which there are 3, 2s and 2, 4s.
Total numbers beginning with 2 = \(\frac{6!}{2!2!}\)
= \(\frac{3 \times 4 \times 5 \times 6}{2}\)
= 180
and total numbers beginning with 4 = \(\frac{6!}{3!}\)
= 4 × 5 × 6
= 120
Therefore, the required number of numbers = 60 + 180 + 120 = 360.
Alternative Method:
The number of 7-digit arrangements, clearly \(\frac{7!}{3!2!}\) = 420.
But this will include those numbers also, which have 0 at the extreme left position.
The number of such arrangements \(\frac{6!}{3!2!}\) (by fixing 0 at the extreme left position) = 60
Therefore the required number of numbers = 420 – 60 = 360
If one or more digits given in the list are repeated, it will be understood that in any number, the digits can be used as many times as is given in the list.
e.g, in the above example, 1 and 0 can be used only once, whereas 2 and 4 can be used 3 times and 2 times, respectively.
Question 24.
In how many ways can 5 girls and 3 boys be seated in a row so that no two boys are together?
Solution:
Let us first seat the 5 girls. This can be done in 5! ways.
For each such arrangement, the three boys can be seated only at the cross-marked places.
× G × G × G × G × G ×
There are 6 cross-marked places, and the three boys can be seated in 6P3 ways.
Hence, by the multiplication principle, the total number of ways = 5! × 6P3
= 5! × \(\frac{6!}{3!}\)
= 4 × 5 × 2 × 3 × 4 × 5 × 6
= 14400
Multiple Choice Questions
Question 1.
A student has 5 pants and 8 shirts. The number of ways in which he can wear the dress in different combinations is
(1) 8P5
(2) 8C5
(3) 8! × 5!
(4) 40
Answer:
(4) 40
Question 2.
If nP4 = 1680, then n is
(1) 7
(2) 8
(3) 9
(4) 10
Answer:
(2) 8
Question 3.
If nPr = 1320, then r is
(1) 3
(2) 4
(3) 5
(4) 6
Answer:
(1) 3
Question 4.
A man has 3 sons and 6 schools within his reach. In how many ways can he send his sons to school if no two of his sons are to read in the same school?
(1) 6P3
(2) 6C3
(3) 63
(4) 36
Answer:
(1) 6P3
Question 5.
The number of arrangements that can be formed by taking all the letters {T, U, E, S, D, A, Y}
(1) 720
(2) 5040
(3) 120
(4) 40320
Answer:
(2) 5040
Question 6.
The number of arrangements that can be made out of the letters in expression a4b3c5 when written in full length is
(1) \(\frac{9!}{3!2!4!}\)
(2) \(\frac{7!}{2!3!2!}\)
(3) \(\frac{5!}{4!3!2!}\)
(4) \(\frac{(12)!}{4!3!5!}\)
Answer:
(4) \(\frac{(12)!}{4!3!5!}\)
Question 7.
The number of permutations of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5, 6 when repetition is allowed is
(1) 6P4
(2) 6C4
(3) 64
(4) 46
Answer:
(3) 64
Question 8.
The number of ways of arranging 8 men and 4 women in a row is
(1) 8!
(2) 4!
(3) (12)!
(4) (11)!
Answer:
(3) (12)!
Question 9.
The number of positive divisors of 1080 is
(1) 8
(2) 16
(3) 32
(4) 40
Answer:
(3) 32
Question 10.
The number of ways of selecting 4 boys and 3 girls from a group of 8 boys and 5 girls is
(1) 8C4 + 5C3
(2) 8C4 × 5C3
(3) 8C3 – 5C3
(4) None
Answer:
(2) 8C4 × 5C3
Question 11.
If 12CS+1 = 12C2S-5, then the value of S is
(1) 4
(2) 8
(3) 12
(4) 6
Answer:
(4) 6
Question 12.
The number of diagonals of a polygon of 10 sides is
(1) 35
(2) 36
(3) 38
(4) 39
Answer:
(1) 35
Question 13.
A polygon has 54 diagonals. Then the number of its sides is
(1) 7
(2) 9
(3) 10
(4) 12
Answer:
(4) 12