Permutation and Combination is a fundamental topic in combinatorics, widely used in JEE Main and Advanced. It deals with counting arrangements and selections without explicitly listing them.
Basic Principle of Counting:
If a task can be performed in \( m \) ways and another independent task in \( n \) ways, then both together can be performed in:
\[
m \times n
\]
This is known as the Fundamental Principle of Counting.
Factorial Notation:
Factorial is defined as:
\[
n! = n \times (n-1) \times (n-2) \times \cdots \times 1
\]
Special case:
\[
0! = 1
\]
Permutations:
A permutation is an arrangement where order matters.
Number of permutations:
\[
{}^nP_r = \frac{n!}{(n-r)!}
\]
Example:
\[
{}^5P_3 = \frac{5!}{2!} = 60
\]
Permutations with Repetition:
If repetition is allowed:
\[
n^r
\]
Circular Permutations:
Number of arrangements in a circle:
\[
(n – 1)!
\]
Combinations:
A combination is a selection where order does not matter.
Number of combinations:
\[
{}^nC_r = \frac{n!}{r!(n-r)!}
\]
Example:
\[
{}^5C_3 = \frac{5!}{3!2!} = 10
\]
Relation Between Permutation and Combination:
\[
{}^nP_r = {}^nC_r \times r!
\]
Important Identities:
1. Symmetry property:
\[
{}^nC_r = {}^nC_{n-r}
\]
2. Pascal’s Identity:
\[
{}^nC_r = {}^{n-1}C_r + {}^{n-1}C_{r-1}
\]
Combinations with Repetition:
\[
{}^{n+r-1}C_r
\]
Question on Permutation
1. How many 4-letter words, with or without meaning, can be formed using the letters of the word ROSE, provided that repetition of letters is not allowed?
A) 12
B) 16
C) 24
D) 48
Answer: C) 24
Explanation: Since all letters in ROSE are distinct, we use permutations (order matters).
Number of ways:
First position: 4 choices
Second position: 3 choices
Third position: 2 choices
Fourth position: 1 choice
Total arrangements:
\(4 \times 3 \times 2 \times 1 = 24\)
Thus, total number of words = \(4! = {}^4P_4 = 24\).
2. There are 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?
A) 6
B) 8
C) 12
D) 16
Answer: C) 12
Explanation:
To generate a signal, we need to choose and arrange 2 flags out of 4 available flags. Since the flags are placed one below the other, the order matters. Therefore, this is a permutation problem.
- Step 1: Total flags = 4
- Step 2: First (top) position → 4 choices
- Step 3: Second (bottom) position → 3 remaining choices
Total number of signals:
$$4 \times 3 = 12$$
Using permutation formula:
$$^4P_2 = \frac{4!}{(4-2)!} = \frac{4 \times 3 \times 2 \times 1}{2 \times 1} = 12$$
Thus, the total number of different signals is 12.
3. How many 2-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, if the digits can be repeated?
A) 6
B) 10
C) 12
D) 25
Answer: B) 10
Explanation:
A 2-digit number has two places: Tens place and Units place. For a number to be even, the digit in the units place must be even.
- Step 1: Available digits → {1, 2, 3, 4, 5}
- Step 2: Units place (even condition)
Only even digits can be used → 2, 4
So, 2 options - Step 3: Tens place
Repetition is allowed, so any digit can be used → 5 options
Total number of 2-digit even numbers:
$$5 \times 2 = 10$$
Possible numbers:
12, 22, 32, 42, 52, 14, 24, 34, 44, 54
Thus, the total number of such numbers is 10.
4. How many different signals 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?
A) 120
B) 320
C) 325
D) 600
Answer: B) 320
Explanation:
“At least 2 flags” means we can use 2, 3, 4, or 5 flags. Since the flags are distinct and arranged in order, this is a permutation problem.
- Case 1: Using 2 flags
$$^5P_2 = 5 \times 4 = 20$$ - Case 2: Using 3 flags
$$^5P_3 = 5 \times 4 \times 3 = 60$$ - Case 3: Using 4 flags
$$^5P_4 = 5 \times 4 \times 3 \times 2 = 120$$ - Case 4: Using 5 flags
$$^5P_5 = 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$$
Total number of signals:
$$20 + 60 + 120 + 120 = 320$$
Thus, the total number of different signals is 320.
5. How many 3-digit numbers can be formed from the digits 1, 2, 3, 4, and 5 if repetition of digits is allowed?
Answer: 125
Explanation:
We are given the digits {1, 2, 3, 4, 5} and need to form 3-digit numbers. Since repetition is allowed, each position can be filled independently.
- Hundreds place: 5 choices
- Tens place: 5 choices
- Units place: 5 choices
Using the Fundamental Principle of Counting:
$$5 \times 5 \times 5 = 125$$
Thus, the total number of 3-digit numbers that can be formed is 125.
6. How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that repetition of the digits is not allowed?
Answer: 60
Explanation:
Since repetition is not allowed, each digit can be used only once.
- Hundreds place: 5 choices (1, 2, 3, 4, 5)
- Tens place: 4 remaining choices
- Units place: 3 remaining choices
Using the Fundamental Principle of Counting:
$$5 \times 4 \times 3 = 60$$
Alternatively, using permutation:
$$^5P_3 = \frac{5!}{(5-3)!} = 60$$
Thus, the total number of 3-digit numbers that can be formed is 60.
7. How many 4-letter codes can be formed using the first 10 letters of the English alphabet (A to J), if no letter can be repeated?
Solution:
- 1st letter → 10 choices
- 2nd letter → 9 choices
- 3rd letter → 8 choices
- 4th letter → 7 choices
Total number of codes:
$$10 \times 9 \times 8 \times 7 = 5040$$
Answer: 5040
8.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: 336
Explanation:
The first two digits are fixed as 6 and 7, so we only need to fill the remaining 3 positions.
- Step 1: Available digits
Total digits = 0 to 9 → 10 digits
Exclude 6 and 7 → Remaining digits = {0, 1, 2, 3, 4, 5, 8, 9} → 8 digits - Step 2: Fill remaining positions
3rd digit → 8 choices
4th digit → 7 choices
5th digit → 6 choices
Using the Fundamental Principle of Counting:
$$1 \times 1 \times 8 \times 7 \times 6 = 336$$
Alternatively, using permutation:
$$^8P_3 = 8 \times 7 \times 6 = 336$$
Thus, the total number of such telephone numbers is 336.
9. A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?
Solution:
Each coin toss has 2 possible outcomes: Head (H) or Tail (T).
- 1st toss → 2 choices
- 2nd toss → 2 choices
- 3rd toss → 2 choices
Total number of outcomes:
$$2 \times 2 \times 2 = 2^3 = 8$$
Possible outcomes:
HHH, HHT, HTH, HTT, THH, THT, TTH, TTT
Answer: 8
10. Find the number of permutations of the letters of the word ALLAHABAD.
Solution:
The word ALLAHABAD has 9 letters.
Frequency of letters:
- A → 4 times
- L → 2 times
- H → 1 time
- B → 1 time
- D → 1 time
Since some letters are repeated, the number of distinct permutations is:
$$\frac{9!}{4! \times 2!}$$
Calculation:
$$9! = 362880,\quad 4! = 24,\quad 2! = 2$$
$$\frac{362880}{24 \times 2} = \frac{362880}{48} = 7560$$
Answer: 7560
11. How many 4-digit numbers can be formed by using the digits 1 to 9, if repetition of digits is not allowed?
Solution:
We have digits from 1 to 9 (total 9 digits), and repetition is not allowed.
- 1st digit → 9 choices
- 2nd digit → 8 choices
- 3rd digit → 7 choices
- 4th digit → 6 choices
Total number of 4-digit numbers:
$$9 \times 8 \times 7 \times 6 = 3024$$
Answer: 3024
12. How many numbers lying between 100 and 1000 can be formed with the digits 0, 1, 2, 3, 4, 5, if repetition of digits is not allowed?
Solution:
Numbers between 100 and 1000 are 3-digit numbers.
Available digits: 0, 1, 2, 3, 4, 5 (total 6 digits)
- Hundreds place: Cannot be 0 → 5 choices (1 to 5)
- Tens place: Remaining 5 choices (including 0, excluding used digit)
- Units place: Remaining 4 choices
Total number of such numbers:
$$5 \times 5 \times 4 = 100$$
Answer: 100
13. Find the value of n such that (i) \( ^nP_5 = 42 \ ^nP_3, \, n > 4 \) (ii) \( \frac{^nP_4}{^{n-1}P_4} = \frac{5}{3}, \, n > 4 \)
(A) n = 10
(B) n = 12
(C) n = 15
(D) n = 8
Answer: (A)
Explanation:
(i) Given \( ^nP_5 = 42 \ ^nP_3 \). Expanding the permutations, we get
\( n(n – 1)(n – 2)(n – 3)(n – 4) = 42 \cdot n(n – 1)(n – 2) \).
Since \( n > 4 \), we divide both sides by \( n(n – 1)(n – 2) \) to get
\( (n – 3)(n – 4) = 42 \).
This simplifies to the quadratic equation
\( n^2 – 7n – 30 = 0 \).
Solving for \( n \), we get
\( (n – 10)(n + 3) = 0 \), giving \( n = 10 \) or \( n = -3 \).
As \( n \) cannot be negative, the value is \( n = 10 \).
(ii) Given \( \frac{^nP_4}{^{n-1}P_4} = \frac{5}{3} \). This can be written as
\( 3 \cdot ^nP_4 = 5 \cdot ^{n-1}P_4 \).
Expanding these gives
\( 3n(n – 1)(n – 2)(n – 3) = 5(n – 1)(n – 2)(n – 3)(n – 4) \).
Dividing by the common terms \( (n – 1)(n – 2)(n – 3) \), we get
\( 3n = 5(n – 4) \).
Solving \( 3n = 5n – 20 \) results in
\( 2n = 20 \), so \( n = 10 \).
In both cases, the value of \( n \) is 10.
14. Find the number of different 8-letter arrangements that can be made from the letters of the word DAUGHTER so that all vowels occur together.
(A) 4320
(B) 720
(C) 4320
(D) 5040
Answer: (A)
Explanation:
All vowels occur together: The word DAUGHTER has 8 distinct letters, including 3 vowels (A, U, E).
To keep vowels together, we treat \( (AUE) \) as a single object.
Thus, we have 6 objects (1 vowel group + 5 consonants), which can be arranged in
\( 6! \) ways.
Within the group, the 3 vowels can be arranged in
\( 3! \) ways.
By the multiplication principle, total arrangements are
\( 6! \times 3! = 720 \times 6 = 4320 \).
15. Find the number of different 8-letter arrangements that can be made from the letters of the word DAUGHTER so that all vowels do not occur together.
(A) 36000
(B) 40320
(C) 35900
(D) 36000
(ii) All vowels do not occur together: Total arrangements of 8 distinct letters are
\( 8! = 40320 \).
Subtracting the cases where vowels are together:
\( 40320 – 4320 = 36000 \).
Hence, the number of arrangements where vowels do not all occur together is 36000.
16. 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?
(A) 1260
(B) 2520
(C) 630
(D) 315
Answer: (A)
Explanation:
The total number of discs to be arranged is
\( 4 + 3 + 2 = 9 \).
Since discs of the same colour are indistinguishable, we use the formula for permutations of objects with repetition:
\( \frac{n!}{p_1! \, p_2! \, p_3!} \),
where \( n \) is the total number of objects and \( p_1, p_2, p_3 \) are the frequencies.
Substituting the values:
\( \frac{9!}{4! \, 3! \, 2!} \)
Calculating:
\( \frac{362880}{24 \times 6 \times 2} = \frac{362880}{288} = 1260 \).
Thus, the number of different arrangements is 1260.
17. Find the number of arrangements of the letters of the word INDEPENDENCE.
Ans: 1663200
Explanation:
The word INDEPENDENCE has 12 letters with repetitions:
3 N’s, 4 E’s, 2 D’s and others distinct.
Total arrangements:
\( \frac{12!}{3! \, 4! \, 2!} = 1663200 \).
(i) Words start with P: Fix P at the beginning.
Remaining 11 letters can be arranged in
\( \frac{11!}{3! \, 4! \, 2!} = 138600 \).
18. In how many of arrangements of the word INDEPENDENCE, do the words start with P?
Ans: 138600
Explanation:
Words start with P: Fix P at the beginning.
Remaining 11 letters can be arranged in
\( \frac{11!}{3! \, 4! \, 2!} = 138600 \).
19. In how many of arrangements of the word INDEPENDENCE, do all vowels occur together?
Ans: 16800
Explanation: All vowels occur together: Vowels are 5 (4 E’s, 1 I).
Treat them as one group. Along with 7 consonants, we have 8 objects.
Arrangements of these objects:
\( \frac{8!}{3! \, 2!} \)
Arrangements of vowels within the group:
\( \frac{5!}{4!} \)
Total arrangements:
\( \frac{8!}{3! \, 2!} \times \frac{5!}{4!} = 16800 \).
20. In how many of arrangements of the word INDEPENDENCE, do all vowels never occur together?
Ans: 1646400
Solution: Vowels never occur together:
\( 1663200 – 16800 = 1646400 \).
21. In how many of arrangements of the word INDEPENDENCE, do the words begin with I and end in P?
Ans: 12600
Solution: Words begin with I and end with P: Fix I at start and P at end.
Remaining 10 letters (3 N’s, 4 E’s, 2 D’s) can be arranged in \( \frac{10!}{3! \, 4! \, 2!} = 12600 \).
22. How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?
(A) 729
(B) 504
(C) 648
(D) 84
Answer: (B)
Explanation: Number of ways = \( ^9P_3 = \frac{9!}{6!} = 9 \times 8 \times 7 = 504 \).
23. How many 4-digit numbers are there with no digit repeated?
(A) 5040
(B) 4536
(C) 3024
(D) 5000
Answer: (B)
Explanation: Thousands place: 9 choices (1–9). Remaining places: \( ^9P_3 = 9 \times 8 \times 7 \).
Total = \( 9 \times 504 = 4536 \).
24. How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?
(A) 60
(B) 40
(C) 120
(D) 30
Answer: (A)
Explanation: Units place: 3 choices (2, 4, 6). Remaining places: \( ^5P_2 = 5 \times 4 = 20 \).
Total = \( 3 \times 20 = 60 \).
25. 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?
(A) 120 total, 48 even
(B) 120 total, 60 even
(C) 24 total, 12 even
(D) 100 total, 40 even
Answer: (A)
Explanation: Total = \( ^5P_4 = 5 \times 4 \times 3 \times 2 = 120 \).
Even numbers: last digit = 2 or 4 → 2 choices. Remaining: \( ^4P_3 = 24 \).
Total even = \( 2 \times 24 = 48 \).
26. From a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman assuming one person cannot hold more than one position?
(A) 64
(B) 56
(C) 28
(D) 16
Answer: (B)
Explanation: Order matters → \( ^8P_2 = 8 \times 7 = 56 \).
27. Find n if \( ^{n-1}P_3 : \ ^nP_4 = 1 : 9 \).
(A) n = 9
(B) n = 10
(C) n = 12
(D) n = 8
Answer: (A)
Explanation:
\( \frac{^{n-1}P_3}{^nP_4} = \frac{1}{9} \Rightarrow \frac{(n-1)(n-2)(n-3)}{n(n-1)(n-2)(n-3)} = \frac{1}{9} \).
So \( \frac{1}{n} = \frac{1}{9} \Rightarrow n = 9 \).
28. Find r if (i) \( ^5P_r = 2 \ ^6P_{r-1} \) (ii) \( ^5P_r = \ ^6P_{r-1} \)
(A) (i) r = 3, (ii) r = 4
(B) (i) r = 3, (ii) r = 9
(C) (i) r = 4, (ii) r = 1
(D) (i) r = 2, (ii) r = 3
Answer: (A)
Explanation:
(i) Leads to equation \( (7-r)(6-r) = 12 \Rightarrow r = 3 \).
(ii) Leads to equation \( (7-r)(6-r) = 6 \Rightarrow r = 4 \).
29. How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once?
(A) 40,320
(B) 36,288
(C) 5,040
(D) 720
Answer: (A) 40,320
Explanation: The word EQUATION consists of 8 distinct letters.
Since each letter must be used exactly once, the number of possible arrangements is given by the permutation of 8 distinct objects.
Using the factorial formula:
\( 8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \)
Calculating:
\( 8! = 40320 \)
Thus, the total number of unique words is 40,320.
30. How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if 4 letters are used at a time?
Answer: 360
Explanation:
The word MONDAY has 6 distinct letters.
4 letters are used at a time:
Number of permutations =
\( ^6P_4 = 6 \times 5 \times 4 \times 3 = 360 \).
30. How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if all letters are used at a time?
Ans:
All letters are used at a time:
Total arrangements =
\( 6! = 720 \).
32. How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if all letters are used but first letter is a vowel?
Ans: 240
All letters are used but first letter is a vowel:
Vowels in MONDAY = O, A → 2 choices for first position.
Remaining 5 letters can be arranged in
\( 5! = 120 \) ways.
Total =
\( 2 \times 120 = 240 \).
33. In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s not come together?
Answer: (B) 33810
Explanation:
The word MISSISSIPPI contains 11 letters with repetitions:
1 M, 4 I’s, 4 S’s, and 2 P’s.
Total number of distinct permutations:
\( \frac{11!}{4! \, 4! \, 2!} = 34650 \).
To find the number of arrangements where the four I’s do not come together, first calculate the cases where they do come together.
Treat the four I’s as a single unit. Then we have 8 objects:
(IIII), M, S, S, S, S, P, P.
Number of such arrangements:
\( \frac{8!}{4! \, 2!} = 840 \).
Hence, required number of arrangements:
\( 34650 – 840 = 33810 \).
Thus, the number of permutations where the four I’s do not come together is 33810.
34. In how many ways can the letters of the word PERMUTATIONS be arranged if words start with P and end with S?
Ans: 1814400
The word PERMUTATIONS has \(12\) letters, with T repeated twice.
Fixing P at the start and S at the end leaves \(10\) letters (including \(2\) Ts) to be arranged, calculated as \( \frac{10!}{2!} = 1814400 \).
35. In how many ways can the letters of the word PERMUTATIONS be arranged if vowels are all together?
Ans: 2419200
Treating the \(5\) vowels (E, U, A, I, O) as one unit, we arrange \(8\) units (the vowel block + \(7\) consonants) in \( \frac{8!}{2!} \) ways and multiply by \(5!\) for the internal vowel arrangements, giving \(2419200\).
36. In how many ways can the letters of the word PERMUTATIONS be arranged if there are always 4 letters between P and S?
Ans: (25401600
There are \(7\) possible positions for the P and S pair (1 and 6, 2 and 7, etc.) and they can swap places (\(2\) ways). The remaining \(10\) letters are arranged in \( \frac{10!}{2!} \) ways. Multiplying \(7 \times 2 \times 1814400\) results in \(25401600\).
Q. The sides \( AB \), \( BC \), and \( CA \) of a triangle \( ABC \) have \( 3 \), \( 5 \), and \( 6 \) interior points respectively on them. The total number of triangles that can be constructed using these points as vertices is:
Solution:
1. Total Points:
\( 3 + 5 + 6 = 14 \) points.
2. Total Selections:
Number of ways to choose any 3 points from 14:
\( ^{14}C_3 = \frac{14 \times 13 \times 12}{3 \times 2 \times 1} = 364 \).
3. Subtract Collinear Cases:
Triangles cannot be formed if all 3 points lie on the same side.
Points on \( AB \): \( ^3C_3 = 1 \)
Points on \( BC \): \( ^5C_3 = 10 \)
Points on \( CA \): \( ^6C_3 = 20 \)
Total collinear cases =
\( 1 + 10 + 20 = 31 \).
4. Final Calculation:
\( 364 – 31 = 333 \).
Thus, the total number of triangles = 333.