Permutations And Combinations

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 × 2 × 1

= 6

= 4 × 3 × 2 × 1

= 24

= 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.

Ans :

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.

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

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 :