**Chapter 6.1: Introduction to Permutations and Combinations**

**Permutation:****Combination:**A selection of objects without regard to order.**Factorial:**The product of all positive integers from 1 to n, denoted as n!.

**Chapter 6.2: Permutations**

**nPr:****Formula for nPr:**nPr = n! / (n – r)!**Permutations with Repetition:**When objects can be repeated, the number of permutations is n^r.

**Chapter 6.3: Combinations**

**nCr:****Formula for nCr:**nCr = n! / (r!(n – r)!)**Combinations with Repetition:**The number of combinations of n objects taken r at a time with repetition allowed is (n + r – 1)Cr.

**Key Concepts:**

- The difference between permutations and combinations
- The formulas for nPr and nCr
- Permutations and combinations with repetition
- Applications of permutations and combinations in various counting problems

**Exercise 6.1**

**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?**

**Ans : **

(i) **Repetition of digits is allowed:**

**Hundreds place:**5 choices (1, 2, 3, 4, or 5)**Tens place:**5 choices (repetition is allowed)**Ones place:**5 choices (repetition is allowed)

Total number of 3-digit numbers = 5 * 5 * 5

= 125

(ii) **Repetition of digits is not allowed:**

**Hundreds place:**5 choices**Tens place:**4 choices (since one digit is already used)**Ones place:**3 choices (since two digits are already used)

Total number of 3-digit numbers = 5 * 4 * 3 = 60

Therefore, there are **125** 3-digit numbers when repetition is allowed and **60** 3-digit numbers when repetition is not allowed.

**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?**

**Ans : **

To form a 3-digit even number, the units place must be filled with an even digit. From the given digits, 2, 4, and 6 are even.

**Units place:** 3 choices (2, 4, or 6)

**Hundreds and tens place:** Since repetition is allowed, each place can be filled with any of the 6 digits (1, 2, 3, 4, 5, or 6).

6 * 6 * 3 = 108

So, there are 108 different 3-digit even numbers that can be formed from the digits 1, 2, 3, 4, 5, and 6 with repetition allowed.

**3. How many 4-letter code can be formed using the first 10 letters of the English alphabet, if no letter can be repeated?**

**Ans : **

The formula for permutations is:

nPr = n! / (n – r)!

Where:

- n is the total number of objects
- r is the number of objects taken at a time

In this case, n = 10 (the first 10 letters of the English alphabet) and r = 4 (the number of letters in the code).

Substituting the values into the formula:

10P4 = 10! / (10 – 4)!

10P4 = 10! / 6!

10P4

= 10 * 9 * 8 * 7

10P4 = 5040

**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?**

**Ans :**

Since the first two digits are fixed as 67, we have 8 remaining digits (0, 1, 2, 3, 4, 5, 8, 9) to fill the last three places.

**Third place:** 8 choices (any of the remaining 8 digits)

**Fourth place:** 7 choices (one digit is already used)

**Fifth place:** 6 choices (two digits are already used)

Total number of 5-digit telephone numbers = 1 * 1 * 8 * 7 * 6 = **336**

**5. A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?**

**Ans : **

A coin has two possible outcomes:

heads or tails.

When a coin is tossed once, there are 2 possible outcomes.

When a coin is tossed twice, there are 2 * 2 = 4 possible outcomes (HH, HT, TH, TT).

When a coin is tossed three times, there are 2 * 2 * 2 = 8 possible outcomes.

**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?**

**Ans : **

The formula for permutations is:

nPr = n! / (n – r)!

Where:

- n is the total number of objects
- r is the number of objects taken at a time

In this case, n = 5 (the number of flags) and r = 2 (the number of flags used for each signal).

Substituting the values into the formula:

5P2 = 5! / (5 – 2)!

5P2 = 5! / 3!

5P2 = 5 * 4 * 3! / 3!

5P2 = 5 * 4

5P2 = 20

Therefore, there are **20** different signals that can be generated using 5 flags of different colors.

**Exercise 6.2**

**1. Evaluate **

**(i) 8 ! (ii) 4 ! – 3 !**

**Ans : **

(i) 8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1

= **40,320**

(ii) 4! – 3! = (4 × 3 × 2 × 1) – (3 × 2 × 1) = 24 – 6 = **18**

**2. Is 3 ! + 4 ! = 7 ! ?**

**Ans : **

**No, 3! + 4! is not equal to 7!**

Let’s calculate each factorial separately:

3!

= 3 × 2 × 1

= 6

4!

= 4 × 3 × 2 × 1

= 24

7!

= 7 × 6 × 5 × 4 × 3 × 2 × 1

= 5040

Now, let’s add 3! and 4!:

3! + 4! = 6 + 24 = 30

As you can see, 30 is not equal to 5040. Therefore, 3! + 4! ≠ 7!.

**3. Compute 8! / 6! * 2!**

**Ans : **

8! = 8 * 7 * 6! 6! = 6 * 5 * 4 * 3 * 2 * 1 2! = 2 * 1

Substituting these values into the expression:

(8 * 7 * 6!) / ((6 * 5 * 4 * 3 * 2 * 1) * (2 * 1))

Simplifying:

(8 * 7 * 6!) / (6! * 2 * 1)

Canceling out the common factor of 6!:

(8 * 7) / (2 * 1)

Simplifying further:

56 / 2

= **28**

Therefore, 8! / (6! * 2!) equals 28.

**4. **

**Ans : **

**5. **

**Ans : **

**(i) n = 6, r = 2**

6P2 = 6! / (6 – 2)!

= 6! / 4! = 6 * 5 * 4! / 4!

= 6 * 5

= 30

**(ii) n = 9, r = 5**

9P5 = 9! / (9 – 5)!

= 9! / 4!

= 9 * 8 * 7 * 6 * 5 * 4! / 4!

= 9 * 8 * 7 * 6 * 5

= 15,120

**Exercise 6.3**

**1. How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?**

**Ans : **

The formula for permutations is:

nPr = n! / (n – r)!

Where:

- n is the total number of objects
- r is the number of objects taken at a time

In this case, n = 9 (the 9 digits) and r = 3 (the number of digits in a 3-digit number).

Substituting the values into the formula:

9P3 = 9! / (9 – 3)!

9P3 = 9! / 6!

9P3 = 9 * 8 * 7 * 6! / 6!

9P3 = 9 * 8 * 7

9P3 = 504

Therefore, there are **504** different 3-digit numbers that can be formed using the digits 1 to 9 if no digit is repeated.

**2. How many 4-digit numbers are there with no digit repeated?**

**Ans : **

The formula for permutations is:

nPr = n! / (n – r)!

Where:

- n is the total number of objects

In this case, n = 10 (the 10 digits) and r = 4 (the number of digits in a 4-digit number).

Substituting the values into the formula:

10P4 = 10! / (10 – 4)!

10P4 = 10! / 6!

10P4

= 10 * 9 * 8 * 7 * 6! / 6!

10P4 = 10 * 9 * 8 * 7

10P4 = 5040

Therefore, there are **5040** different 4-digit numbers that can be formed using the digits 0 to 9 if no digit is repeated.

**3. How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?**

**Ans : **

To form a 3-digit even number, the units place must be filled with an even digit.

**Units place:** 3 choices (2, 4, or 6)

**Hundreds and tens place:** Since repetition is not allowed, we have 5 choices for the hundreds place and 4 choices for the tens place.

3 * 5 * 4 = 60

So, there are 60 different 3-digit even numbers that can be formed using the digits 1, 2, 3, 4, 6, and 7 with no digit repeated.

**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?**

**Ans : **

To find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, and 5 without repetition, we can use the concept of permutations.

**Number of 4-digit numbers:**

Since we have 5 digits and we want to choose 4 of them, we can use the permutation formula:

nPr = n! / (n – r)!

where n is the total number of objects (5 digits in this case) and r is the number of objects we want to choose (4 digits in this case).

Substituting the values into the formula:

5P4 = 5! / (5 – 4)! 5P4 = 5! / 1! 5P4 = 5 * 4 * 3 * 2 * 1 5P4 = 120

Therefore, there are **120** different 4-digit numbers that can be formed using the digits 1, 2, 3, 4, and 5 without repetition.

**Number of even 4-digit numbers:**

For a number to be even, its units digit must be even. From the given digits, only 2 and 4 are even. So, we can fix the units place with either 2 or 4.

Now, we have 4 digits left (1, 3, 5, and either 2 or 4) to fill the remaining three places. We can use permutations again:

4P3 = 4! / (4 – 3)! 4P3 = 4! / 1! 4P3 = 4 * 3 * 2 * 1 4P3 = 24

Since we can choose either 2 or 4 for the units place, the total number of even 4-digit numbers is:

2 * 24 = **48**

**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?**

**Ans : **

**Choosing the chairman:**

There are 8 people to choose from, so there are 8 ways to choose the chairman.

**Choosing the vice chairman:**

Once the chairman is chosen, there are 7 people left to choose from for the vice chairman.

**Total number of ways:**

To find the total number of ways to choose both the chairman and the vice chairman, we multiply the number of ways to choose each position:

8 * 7 = 56

Therefore, there are **56** ways to choose a chairman and a vice chairman from a committee of 8 people assuming one person cannot hold more than one position.

**6. Find n if n – 1P3 : nP4 = 1 : 9.**

**Ans : **

We are given the ratio:

n – 1P3 : nP4

= 1 : 9

We can write this as a fraction:

(n – 1P3) / (nP4) = 1/9

Now, let’s substitute the formulas for n – 1P3 and nP4:

((n – 1)! / (n – 1 – 3)!) / (n! / (n – 4)!) = 1/9

Simplifying the expression:

((n – 1)! / (n – 4)!) / (n * (n – 1) * (n – 2) * (n – 3) * (n – 4)!) = 1/9

Canceling out common factors:

1 / (n * (n – 2) * (n – 3)) = 1/9

Therefore, we have:

n * (n – 2) * (n – 3) = 9

Expanding the left side:

n³ – 5n² + 6n = 9

Rearranging the equation:

n³ – 5n² + 6n – 9 = 0

We can solve this cubic equation using a calculator or computer software. The solution is:

n = 9

Therefore, the value of n is **9**.

**7. **

**Ans : **

**8. How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once?**

**Ans : **

The formula for permutations is:

nPr = n! / (n – r)!

Where:

- n is the total number of objects

In this case, n = 8 (the 8 letters) and r = 8 (since we want to use all the letters).

Substituting the values into the formula:

8P8 = 8! / (8 – 8)!

8P8 = 8! / 0!

Note that 0! is defined as 1. So, we have:

8P8 = 8! / 1

8P8

= 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1

8P8 = 40,320

Therefore, there are **40,320** different words, with or without meaning, that can be formed using all the letters of the word EQUATION, using each letter exactly once.

**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?**

**Ans : **

**(i) **

We can use the permutation formula:

nPr = n! / (n – r)!

Where:

- n is the total number of objects
- r is the number of objects taken at a time

In this case, n = 6 (the 6 letters in MONDAY) and r = 4 (the number of letters used).

Substituting the values:

6P4 = 6! / (6 – 4)!

6P4 = 6! / 2!

6P4 = 6 * 5 * 4 * 3 * 2! / 2!

6P4 = 360

Therefore, there are **360** different 4-letter words that can be formed.

**(ii) **

Since we’re using all 6 letters, this is the same as finding the number of permutations of 6 objects taken 6 at a time:

6P6 = 6! / (6 – 6)!

6P6 = 6! / 0!

6P6 = 6 * 5 * 4 * 3 * 2 * 1

6P6 = 720

Therefore, there are **720** different 6-letter words (all letters used).

**(iii)**

There are 2 vowels in MONDAY (O and A). So, the first letter can be chosen in 2 ways.

For the remaining 5 letters, we have 5 * 4 * 3 * 2 * 1 = 120 ways to arrange them.

Total number of words = 2 * 120 = 240

**10. In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s not come together? **

**Ans : **

First, let’s find the total number of distinct permutations of the letters in MISSISSIPPI.

- There are 11 letters in total.
- The letter ‘I’ appears 4 times.
- The letter ‘S’ appears 4 times.
- The letter ‘P’ appears 2 times.

Using the formula for permutations with repetitions:

Total permutations = 11! / (4! * 4! * 2!)

**Permutations with all I’s together:**

To find the number of permutations where all four I’s are together, we can consider the group of four I’s as a single entity. So, we have 8 entities (the group of I’s, 4 S’s, and 2 P’s).

Permutations with all I’s together = 8! / (4! * 2!)

**Permutations with I’s not together:**

The number of permutations with I’s not together is the total permutations minus the permutations with all I’s together.

Permutations with I’s not together = Total permutations – Permutations with all I’s together

= (11! / (4! * 4! * 2!)) – (8! / (4! * 2!))

**Calculating the values:**

- 11! / (4! * 4! * 2!) ≈ 34,650
- 8! / (4! * 2!) = 420

Therefore, the number of distinct permutations of the letters in MISSISSIPPI where the four I’s do not come together is approximately:

34,650 – 420 = **34,230**

**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? **

**Ans : **

**Analyzing the Word “PERMUTATIONS”**

**Total Letters:** 11 **Vowels:** E, U, A, I, O (5 vowels) **Consonants:** P, R, M, T, T, N, S (7 consonants)

**(i) Words start with P and end with S**

**Fixed Positions:**P and S are fixed.**Remaining Letters:**9 (E, R, M, U, T, T, A, I, O)**Arrangements:**9! / 2! (due to 2 T’s)

**Number of words:** 1 * 9! / 2! = 181,440

**(ii) Vowels are all together**

**Treat Vowels as a Single Unit:**Consider “EUAIO” as a single unit.**Total Units:**7 (1 unit of vowels, 6 consonants)**Arrangements:**7! * 5! (5! for arranging vowels within the unit)

**Number of words:** 7! * 5! = 24,19,200

**(iii) There are always 4 letters between P and S**

**Possible Positions for P and S:**- (1, 6), (2, 7), (3, 8), (4, 9), (5, 10), (6, 11), (7, 12)
**Total Positions:**7**Arrangements for Remaining Letters:**10! / 2! (due to 2 T’s)

**Number of words:** 7 * 10! / 2! = 25,40,1600

**Exercise 6.4**

**1. If nC8 = nC2 , find nC2 .**

**Ans : **

**2. Determine n if (i) 2nC3 : nC3 = 12 : 1 (ii) 2nC3 : nC3 = 11 : 1**

**Ans : **

**3. How many chords can be drawn through 21 points on a circle?**

**Ans : **

The formula for combinations is:

nCr = n! / (r!(n – r)!)

Where:

- n is the total number of objects (21 points in this case)
- r is the number of objects we want to choose (2 points in this case)

Substituting the values into the formula:

21C2 = 21! / (2!(21 – 2)!)

21C2 = 21! / (2! * 19!)

21C2 = 21 * 20 * 19! / (2 * 1 * 19!)

21C2 = 21 * 20 / 2

21C2 = 210

Therefore, there are **210** different chords that can be drawn through 21 points on a circle.

**4. In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls? **

**Ans : **

nCr = n! / (r!(n – r)!)

First, we need to find the number of ways to choose 3 boys from 5 boys:

5C3 = 5! / (3!(5 – 3)!)

5C3 = 5! / (3! * 2!)

5C3 = 5 * 4 * 3! / (3! * 2!)

5C3 = 5 * 4 / 2

5C3 = 10

Next, we need to find the number of ways to choose 3 girls from 4 girls:

4C3 = 4! / (3!(4 – 3)!)

4C3 = 4! / (3! * 1!)

4C3 = 4 * 3! / 3!

4C3 = 4

Finally, to find the total number of ways to choose 3 boys and 3 girls, we multiply the number of ways to choose boys by the number of ways to choose girls:

Total number of ways = 5C3 * 4C3 = 10 * 4 = 40

Therefore, there are **40** ways to select a team of 3 boys and 3 girls from 5 boys and 4 girls.

**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**

**Ans : **

First, we need to find the number of ways to choose 3 red balls from 6 red balls:

6C3 = 6! / (3!(6 – 3)!)

6C3 = 6! / (3! * 3!)

6C3 =

6 * 5 * 4 * 3! / (3! * 3!)

6C3 = 6 * 5 * 4 / 3!

6C3 = 20

Similarly, we can find the number of ways to choose 3 white balls from 5 white balls and 3 blue balls from 5 blue balls:

5C3 = 5! / (3!(5 – 3)!) = 10

5C3 = 5! / (3!(5 – 3)!) = 10

Finally, to find the total number of ways to select 9 balls with 3 of each color, we multiply the number of ways to choose each color:

Total number of ways = 6C3 * 5C3 * 5C3 = 20 * 10 * 10 = 2000

**6. Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination**

**Ans : **

There are 52 – 4 = 48 non-ace cards.

We need to choose 4 of them. So, we can do this in 48C4 ways.

**Total combinations:**

To find the total number of combinations, we multiply the number of ways to choose one ace and the number of ways to choose four non-ace cards:

Total combinations = 4C1 * 48C4

Using the combination formula:

nCr = n! / (r!(n – r)!)

We can calculate the values:

4C1 = 4! / (1!(4 – 1)!) = 4 48C4 = 48! / (4!(48 – 4)!) = 194,580

4 * 194,580 = **778,320**

**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?**

**Ans : **

**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.**

**Ans : **

nCr = n! / (r!(n – r)!)

First, we need to find the number of ways to choose 2 black balls from 5 black balls:

5C2 = 5! / (2!(5 – 2)!)

5C2 = 5! / (2! * 3!)

5C2 = 5 * 4 * 3! / (3! * 2!)

5C2 = 5 * 4 / 2

5C2 = 10

Next, we need to find the number of ways to choose 3 red balls from 6 red balls:

6C3 = 6! / (3!(6 – 3)!)

6C3 = 6! / (3! * 3!)

6C3 = 6 * 5 * 4 * 3! / (3! * 3!)

6C3 = 6 * 5 * 4 / 3!

6C3 = 20

Finally, to find the total number of ways to choose 2 black and 3 red balls, we multiply the number of ways to choose black balls by the number of ways to choose red balls:

Total number of ways = 5C2 * 6C3 = 10 * 20 = 200

**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? **

**Ans : **