# Chapter – 7: Permutations and Combinations

# Exercise 7.1

**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?**

**Solution: –**

(i) Let the 3-digit number be ABC, where C is at the units place, B at the tens place and A at the hundreds place.

Now when repetition is allowed,

The number of digits possible at C is 5. As repetition is allowed, the number of digits possible at B and A is also 5 at each.

Hence, the total number possible 3-digit numbers =5 × 5 × 5 =125

(ii) Let the 3-digit number be ABC, where C is at the units place, B at the tens place and A at the hundreds place.

Now when repetition is not allowed,

The number of digits possible at C is 5. Let’s suppose one of 5 digits occupies place C, now as the repletion is not allowed, the possible digits for place B are 4 and similarly there are only 3 possible digits for place A.

Therefore, The total number of possible 3-digit numbers=5 × 4 × 3=60

**2. How many 3-digits even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?**

**Solution: –**

Let the 3-digit number be ABC, where C is at the unit’s place, B at the tens place and A at the hundreds place.

As the number has to even, the digits possible at C are 2 or 4 or 6. That is number of possible digits at C is 3.

Now, as the repetition is allowed, the digits possible at B is 6. Similarly, at A, also, the number of digits possible is 6.

Therefore, The total number possible 3 digit numbers = 6 × 6 × 3 = 108.

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

**Solution: –**

Let the 4 digit code be 1234.

At the first place, the number of letters possible is 10.

Let’s suppose any 1 of the ten occupies place 1.

Now, as the repetition is not allowed, the number of letters possible at place 2 is 9. Now at 1 and 2, any 2 of the 10 alphabets have been taken. The number of alphabets left for place 3 is 8 and similarly the number of alphabets possible at 4 is 7.

Therefore the total number of 4 letter codes=10 × 9 × 8 × 7=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?**

**Solution: –**

Let the five-digit number be ABCDE. Given that first 2 digits of each number is 67. Therefore, the number is 67CDE.

As the repetition is not allowed and 6 and 7 are already taken, the digits available for place C are 0,1,2,3,4,5,8,9. The number of possible digits at place C is 8. Suppose one of them is taken at C, now the digits possible at place D is 7. And similarly, at E the possible digits are 6.

∴The total five-digit numbers with given conditions = 8 × 7 × 6 = 336.

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

**Solution: –**

Given A coin is tossed 3 times and the outcomes are recorded

The possible outcomes after a coin toss are head and tail.

The number of possible outcomes at each coin toss is 2.

∴The total number of possible outcomes after 3 times = 2 × 2 × 2 = 8.

**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?**

**Solution: –**

Given 5 flags of different colours

We know the signal requires 2 flags.

The number of flags possible for upper flag is 5.

Now as one of the flag is taken, the number of flags remaining for lower flag in the signal is 4.

The number of ways in which signal can be given = 5 × 4 = 20.

# Exercise 7.2

**1. Evaluate(i) 8!**

**(ii) 4! – 3!**

**Solution: –**

(i) Consider 8!

We know that 8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1

= 40320

(ii) Consider 4!-3!

4!-3! = (4 × 3!) – 3!

Above equation can be written as

= 3! (4-1)

= 3 × 2 × 1 × 3

= 18

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

**Solution: –**

Consider LHS 3! + 4!

Computing left hand side, we get

3! + 4! = (3 × 2 × 1) + (4 × 3 × 2 × 1)

= 6 + 24

= 30

Again consider RHS and computing we get

7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040

Therefore LHS ≠ RHS

Therefore 3! + 4! ≠ 7!

**3. Compute **

**Solution: –**

**4.** **If **

**find x.**

**Solution: –**

**5.** **Evaluate**

**,**

**When(i) n = 6, r = 2(ii) n = 9, r = 5**

**Solution: –**

# Exercise 7.3

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

**Solution: –**

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

**Solution: –**

To find four digit number (digits does not repeat)

Now we will have 4 places where 4 digits are to be put.

So, at thousand’s place = There are 9 ways as 0 cannot be at thousand’s place = 9 ways

At hundredth’s place = There are 9 digits to be filled as 1 digit is already taken = 9 ways

At ten’s place = There are now 8 digits to be filled as 2 digits are already taken = 8 ways

At unit’s place = There are 7 digits that can be filled = 7 ways

Total Number of ways to fill the four places = 9 × 9 × 8 × 7 = 4536 ways.

So a total of 4536 four digit numbers can be there with no digits 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?**

**Solution: –**

Even number means that last digit should be even,

Number of possible digits at one’s place = 3 (2, 4 and 6)

⇒ Number of permutations=

One of digit is taken at one’s place, Number of possible digits available = 5

⇒ Number of permutations=

Therefore, total number of permutations =3 × 20=60.

**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?**

**Solution: –**

Total number of digits possible for choosing = 5

Number of places for which a digit has to be taken = 4

As there is no repetition allowed,

⇒ Number of permutations =

The number will be even when 2 and 4 are at one’s place.

The possibility of (2, 4) at one’s place = 2/5 = 0.4

Total number of even number = 120 × 0.4 = 48.

**5.** **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?**

**Solution: –**

Total number of people in committee = 8

Number of positions to be filled = 2

⇒ Number of permutations =

**6.** **Find n if **^{n-1}**P**_{3}**: **^{n}**P**_{3}** = 1: 9.**

**Solution: –**

**7. Find r if**

**(i)**^{5}**P**_{r}** = 2**^{6}**P**_{r-1}** **

**(ii) **^{5}**P**_{r}** = **^{6}**P**_{r-1}

**Solution: –**

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

**Solution: –**

Total number of different letters in EQUATION = 8

Number of letters to be used to form a word = 8

⇒ Number of permutations =

**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?**

**Solution: –**

(i) Number of letters to be used =4

⇒ Number of permutations =

(ii) Number of letters to be used = 6

⇒ Number of permutations =

(iii) Number of vowels in MONDAY = 2 (O and A)

⇒ Number of permutations in vowel =

Now, remaining places = 5

Remaining letters to be used =5

⇒ Number of permutations =

Therefore, total number of permutations = 2 × 120 =240.

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

**Solution: –**

Total number of letters in MISSISSIPPI =11

Letter Number of occurrence

M | 1 |

I | 4 |

S | 4 |

P | 2 |

⇒ Number of permutations =

We take that 4 I’s come together, and they are treated as 1 letter,

∴ Total number of letters=11 – 4 + 1 = 8

⇒ Number of permutations =

Therefore, total number of permutations where four I’s don’t come together = 34650-840=33810

**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?**

**Solution: –**

(i) Total number of letters in PERMUTATIONS =12

Only repeated letter is T; 2times

First and last letter of the word are fixed as P and S respectively.

Number of letters remaining =12 – 2 = 10

⇒ Number of permutations =

(ii) Number of vowels in PERMUTATIONS = 5 (E, U, A, I, O)

Now, we consider all the vowels together as one.

Number of permutations of vowels = 120

Now total number of letters = 12 – 5 + 1= 8

⇒ Number of permutations =

Therefore, total number of permutations = 120 × 20160 = 2419200

(iii) Number of places are as 1 2 3 4 5 6 7 8 9 10 11 12

There should always be 4 letters between P and S.

Possible places of P and S are 1 and 6, 2and 7, 3 and 8, 4 and 9, 5 and 10, 6 and 11, 7 and 12

Possible ways =7,

Also, P and S can be interchanged,

No. of permutations =2 × 7 =14

Remaining 10 places can be filled with 10 remaining letters,

∴ No. of permutations =

Therefore, total number of permutations = 14 × 1814400 =25401600.

# Exercise 7.4

**1. If **^{n}**C**_{8}** = **^{n}**C**_{2}**, find **^{n}**C**_{2.}

**Solution: –**

**2. Determine n if(i) **

^{2n}**C**

_{3:}

^{n}**C**

_{3}**= 12: 1**

(ii)

(ii)

^{2n}**C**

_{3}**:**

^{n}**C**

_{3}**= 11: 1**

**Solution: –**

Simplifying and computing

⇒ 4 × (2n – 1) = 12 × (n – 2)

⇒ 8n – 4 = 12n – 24

⇒ 12n – 8n = 24 – 4

⇒ 4n = 20

∴ n = 5

⇒ 11n – 8n = 22 – 4

⇒ 3n = 18

∴ n = 6

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

**Solution: –**

Given 21 points on a circle

We know that we require two points on the circle to draw a chord

∴ Number of chords is are

⇒ ^{21}C_{2}=

∴ Total number of chords can be drawn are 210

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

**Solution: –**

Given 5 boys and 4 girls are in total

We can select 3 boys from 5 boys in ^{5}C_{3} ways

Similarly, we can select 3 boys from 54 girls in ^{4}C_{3} ways

∴ Number of ways a team of 3 boys and 3 girls can be selected is ^{5}C_{3} × ^{4}C_{3}

⇒ ^{5}C_{3} × ^{4}C_{3} =

⇒ ^{5}C_{3} × ^{4}C_{3} = 10 × 4 = 40

∴ Number of ways a team of 3 boys and 3 girls can be selected is ^{5}C_{3} × ^{4}C_{3} = 40 ways

**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.**

**Solution: –**

Given 6 red balls, 5 white balls and 5 blue balls

We can select 3 red balls from 6 red balls in ^{6}C_{3} ways

Similarly, we can select 3 white balls from 5 white balls in ^{5}C_{3} ways

Similarly, we can select 3 blue balls from 5 blue balls in ^{5}C_{3} ways

∴ Number of ways of selecting 9 balls is ^{6}C_{3} ×^{5}C_{3} × ^{5}C_{3}

∴ 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 is ^{6}C_{3} ×^{5}C_{3} × ^{5}C_{3} = 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.**

**Solution: –**

Given a deck of 52 cards

There are 4 Ace cards in a deck of 52 cards.

According to question, we need to select 1 Ace card out the 4 Ace cards

∴ Number of ways to select 1 Ace from 4 Ace cards is ^{4}C_{1}

⇒ More 4 cards are to be selected now from 48 cards (52 cards – 4 Ace cards)

∴ Number of ways to select 4 cards from 48 cards is ^{48}C_{4}

∴ Number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination 778320.

**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?**

**Solution: –**

Given 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers

There are 5 players how bowl, and we can require 4 bowlers in a team of 11

∴ Number of ways in which bowlers can be selected are: ^{5}C_{4}

Now other players left are = 17 – 5(bowlers) = 12

Since we need 11 players in a team and already 4 bowlers are selected, we need to select 7 more players from 12.

∴ Number of ways we can select these players are: ^{12}C_{7}

∴ Total number of combinations possible are: ^{5}C_{4} × ^{12}C_{7}

∴ Number of ways we can select a team of 11 players where 4 players are bowlers from 17 players are 3960

**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.**

**Solution: –**

Given a bag contains 5 black and 6 red balls

Number of ways we can select 2 black balls from 5 black balls are ^{5}C_{2}

Number of ways we can select 3 red balls from 6 red balls are ^{6}C_{3}

Number of ways 2 black and 3 red balls can be selected are ^{5}C_{2}× ^{6}C_{3}

∴ Number of ways in which 2 black and 3 red balls can be selected from 5 black and 6 red balls are 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?**

**Solution: –**

Given 9 courses are available and 2 specific courses are compulsory for every student

Here 2 courses are compulsory out of 9 courses, so a student need to select 5 – 2 = 3 courses

∴ Number of ways in which 3 ways can be selected from 9 – 2(compulsory courses) = 7 are ^{7}C_{3}

∴ Number of ways a student selects 5 courses from 9 courses where 2 specific courses are compulsory are: 35

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