Complete Exercise 1.2 Class 12 Solutions: Permutation & Combination.

Class 12 Mathematics

Exercise 1.2 - Permutations and Combinations

Table of Contents

• Problem 1: Permutations of 5 objects taken 3 at a time
• Problem 2: Three persons on ten vacant seats
• Problem 3: Vehicle plates and digit arrangements
• Problem 4: Arranging boys and girls
• Problem 5: Eight people seating arrangements
• Problem 6: Six books arrangement
• Problem 7: Beads arrangement
• Problem 8: Word arrangements
• Problem 9: 6-digit numbers from given digits
• Problem 10: Circular arrangements
• Problem 11: Round table seating
• Problem 12: Bracelet arrangements
• Problem 13: Multiple counting problems
• Problem 14: MONDAY arrangements
• Problem 15: COLLEGE and ARRANGE arrangements
• Problem 16: COMPUTER arrangements
• Problem 17: Laptop arrangements
• Problem 18: Internet arrangements

Solutions

Problem 1

Find the number of permutations of five different objects taken three at a time.

Solution

We use the permutation formula:

P(n, r) = n! / (n - r)!

Here, n = 5, r = 3

P(5, 3) = 5! / (5 - 3)! = 5! / 2! = (5 × 4 × 3 × 2 × 1) / (2 × 1) = 5 × 4 × 3 = 60

Answer: 60

Problem 2

If three persons enter a bus in which there are ten vacant seats, find in how many ways they can sit?

Solution

This is a permutation problem where we're selecting 3 seats out of 10 and arranging 3 people in them.

P(10, 3) = 10! / (10 - 3)! = 10! / 7! = 10 × 9 × 8 = 720

Answer: 720

Problem 3

a) How many plates of vehicles consisting of 4 different digits can be made out of the integers 4, 5, 6, 7, 8, 9? How many of these numbers are divisible by 2?

b) How many numbers of 4 different digit numbers can be formed from the digits 2, 3, 4, 5, 6, 7? How many of these numbers are i) divisible by 5? ii) not divisible by 5?

c) How many 5 digit odd numbers can be formed using the digits 3, 4, 5, 6, 7, 8 and 9. If i) repetition of digits is not allowed? ii) repetition of digits is allowed?

Solution

Part (a)

Total 4-digit numbers with different digits from {4,5,6,7,8,9}:

P(6, 4) = 6! / (6-4)! = 6! / 2! = 6 × 5 × 4 × 3 = 360

Numbers divisible by 2 must end with an even digit. Even digits available: 4,6,8

Fix the last digit (3 choices), then arrange the remaining 3 digits from the remaining 5 digits:

3 × P(5, 3) = 3 × (5 × 4 × 3) = 3 × 60 = 180

Part (b)

Total 4-digit numbers with different digits from {2,3,4,5,6,7}:

P(6, 4) = 6! / 2! = 6 × 5 × 4 × 3 = 360

i) Divisible by 5: Must end with 5 (only option)

Fix last digit as 5, then arrange remaining 3 digits from 5 digits:

1 × P(5, 3) = 1 × (5 × 4 × 3) = 60

ii) Not divisible by 5: Total - Divisible by 5 = 360 - 60 = 300

Part (c)

Digits: 3,4,5,6,7,8,9 (7 digits)

i) No repetition allowed, 5-digit odd numbers:

Odd digits available: 3,5,7,9 (4 choices for last digit)

Fix last digit (4 choices), then arrange remaining 4 digits from remaining 6 digits:

4 × P(6, 4) = 4 × (6 × 5 × 4 × 3) = 4 × 360 = 1440

ii) Repetition allowed, 5-digit odd numbers:

First digit: Cannot be 0, so 7 choices

Second, third, fourth digits: 7 choices each

Last digit: Must be odd (3,5,7,9) - 4 choices

7 × 7 × 7 × 7 × 4 = 7⁴ × 4 = 2401 × 4 = 9604

Answers: (a) 360, 180; (b) 360, 60, 300; (c) 1440, 9604

Problem 4

In how many ways can four boys and three girls be seated in a row containing seven seats

a) if they may sit anywhere?

b) if the boys and girls must alternate?

c) if all three girls are together?

d) if girls are to occupy odd seats?

Solution

Part (a)

All 7 persons are distinct, so:

7! = 5040

Part (b)

Boys and girls alternate. With 4 boys and 3 girls, pattern must be: B G B G B G B

Arrange 4 boys: 4! ways

Arrange 3 girls: 3! ways

4! × 3! = 24 × 6 = 144

Part (c)

Treat the 3 girls as one block. Now we have 4 boys + 1 block = 5 entities to arrange.

Arrange 5 entities: 5! ways

Arrange 3 girls within the block: 3! ways

5! × 3! = 120 × 6 = 720

Part (d)

Odd seats: 1,3,5,7 (4 seats)

Even seats: 2,4,6 (3 seats)

Girls must occupy odd seats: Select 3 odd seats from 4 for girls: C(4,3) = 4 ways

Arrange 3 girls in these seats: 3! ways

Arrange 4 boys in remaining 4 seats: 4! ways

C(4,3) × 3! × 4! = 4 × 6 × 24 = 576

Answers: (a) 5040; (b) 144; (c) 720; (d) 576

Problem 5

In how many ways can eight people be seated in a row of eight seats so that two particular persons are

a) always together?

b) never together?

Solution

Part (a)

Treat the two particular persons as one block. Now we have 7 entities to arrange.

Arrange 7 entities: 7! ways

Arrange 2 persons within the block: 2! ways

7! × 2! = 5040 × 2 = 10080

Part (b)

Total arrangements without restrictions: 8! = 40320

Arrangements where they are together: 10080 (from part a)

Arrangements where they are never together: Total - Together

40320 - 10080 = 30240

Answers: (a) 10080; (b) 30240

Problem 6

Six different books are arranged on a shelf. Find the number of different ways in which the two particular books are

a) always together

b) not together

Solution

Part (a)

Treat the two particular books as one block. Now we have 5 entities to arrange.

Arrange 5 entities: 5! ways

Arrange 2 books within the block: 2! ways

5! × 2! = 120 × 2 = 240

Part (b)

Total arrangements without restrictions: 6! = 720

Arrangements where they are together: 240 (from part a)

Arrangements where they are not together: Total - Together

720 - 240 = 480

Answers: (a) 240; (b) 480

Problem 7

In how many ways can four red beads, five white beads and three blue beads be arranged in a row?

Solution

Total beads = 4 + 5 + 3 = 12 beads

But there are identical beads, so we use the formula for permutations of multiset:

12! / (4! × 5! × 3!)

Let's calculate:

12! = 479001600

4! = 24, 5! = 120, 3! = 6

Denominator = 24 × 120 × 6 = 17280

479001600 / 17280 = 27720

Answer: 27720

Problem 8

In how many ways can the letters of the following words be arranged?

a) ELEMENT

b) NOTATION

c) MATHEMATICS

d) MISSISSIPPI

Solution

Part (a) ELEMENT

Total letters: 7

E appears 3 times, L, M, N, T each appear once

7! / 3! = 5040 / 6 = 840

Part (b) NOTATION

Total letters: 8

N appears 2 times, O appears 2 times, T appears 2 times, A and I appear once each

8! / (2! × 2! × 2!) = 40320 / 8 = 5040

Part (c) MATHEMATICS

Total letters: 11

M appears 2 times, A appears 2 times, T appears 2 times, H, E, I, C, S appear once each

11! / (2! × 2! × 2!) = 39916800 / 8 = 4989600

Part (d) MISSISSIPPI

Total letters: 11

M appears 1 time, I appears 4 times, S appears 4 times, P appears 2 times

11! / (4! × 4! × 2!) = 39916800 / (24 × 24 × 2) = 39916800 / 1152 = 34650

Answers: (a) 840; (b) 5040; (c) 4989600; (d) 34650

Problem 9

How many numbers of 6 digits can be formed with the digits 2, 3, 2, 0, 3, 3?

Solution

We have digits: 2, 3, 2, 0, 3, 3

Frequency: 2 appears 2 times, 3 appears 3 times, 0 appears 1 time

Total arrangements of all 6 digits: 6! / (2! × 3!) = 720 / (2 × 6) = 720 / 12 = 60

But we cannot have 0 as the first digit (that would make it a 5-digit number).

Number of arrangements with 0 as first digit: Fix 0 at first position, arrange remaining 5 digits (2,3,2,3,3):

5! / (2! × 3!) = 120 / (2 × 6) = 120 / 12 = 10

Valid 6-digit numbers: Total arrangements - Invalid arrangements

60 - 10 = 50

Answer: 50

Problem 10

In how many ways can 4 Art students and 4 Science students be arranged in a circular table if

a) they may sit anywhere?

b) they sit alternately?

Solution

Part (a)

Total 8 persons arranged around a circular table:

(8 - 1)! = 7! = 5040

Part (b)

Art and Science students alternate. With 4 of each, pattern is fixed: A S A S A S A S

Fix one Art student to account for circular permutations: (4-1)! = 3! ways to arrange remaining Art students

Arrange 4 Science students: 4! ways

3! × 4! = 6 × 24 = 144

Answers: (a) 5040; (b) 144

Problem 11

In how many ways can eight people be seated in a round table if two people insist in sitting next to each other?

Solution

Treat the two particular persons as one block. Now we have 7 entities to arrange around a circular table.

Number of circular arrangements of 7 entities: (7-1)! = 6! = 720

Arrange 2 persons within the block: 2! = 2 ways

720 × 2 = 1440

Answer: 1440

Problem 12

In how many ways can seven different coloured beads be made into a bracelet?

Solution

For a bracelet (which can be flipped), the number of distinct arrangements is:

(n-1)! / 2

For n = 7:

(7-1)! / 2 = 6! / 2 = 720 / 2 = 360

Answer: 360

Problem 13

a) In how many ways can 4 letters be posted in six letter boxes?

b) How many even numbers of 3 digits can be formed when repetition of digits is allowed?

c) In how many ways can 3 prizes be distributed among 4 students so that each student may receive any number of prizes?

Solution

Part (a)

Each of the 4 letters can go into any of the 6 boxes:

6 × 6 × 6 × 6 = 6⁴ = 1296

Part (b)

3-digit even numbers with repetition allowed:

Last digit must be even: 0,2,4,6,8 (5 choices)

First digit cannot be 0: 9 choices (1-9)

Second digit: 10 choices (0-9)

9 × 10 × 5 = 450

Part (c)

Each of the 3 prizes can be given to any of the 4 students:

4 × 4 × 4 = 4³ = 64

Answers: (a) 1296; (b) 450; (c) 64

Problem 14

In how many ways can the letters of the word "MONDAY" be arranged? How many of these arrangements do not begin with M? How many begin with M and do not end with Y?

Solution

Total arrangements:

MONDAY has 6 distinct letters:

6! = 720

Arrangements that do not begin with M:

Total arrangements - Arrangements beginning with M

Arrangements beginning with M: Fix M at first position, arrange remaining 5 letters: 5! = 120

720 - 120 = 600

Arrangements beginning with M and not ending with Y:

Fix M at first position (1 way)

Last position cannot be Y, so we have 4 choices for last position (O,N,D,A)

Arrange remaining 4 letters in the middle 4 positions: 4! = 24

1 × 4 × 24 = 96

Answers: 720, 600, 96

Problem 15

Show that the number of ways in which the letters of the word

a) "COLLEGE" can be arranged so that the two E's always come together is 360.

b) "ARRANGE" can be arranged so that no two R's come together is 900.

Solution

Part (a) COLLEGE

Total letters: 7, with E appearing 2 times, L appearing 2 times, C, O, G appearing once each

Treat the two E's as one block. Now we have 6 entities: EE, C, O, L, L, G

But L appears twice, so:

6! / 2! = 720 / 2 = 360

This proves the statement.

Part (b) ARRANGE

Total letters: 7, with A appearing 2 times, R appearing 2 times, N, G, E appearing once each

First, arrange the letters without R's: A, A, N, G, E

Number of arrangements: 5! / 2! = 120 / 2 = 60

These create 6 gaps where we can place the R's: _ X _ X _ X _ X _ X _

We need to place 2 R's in these 6 gaps: C(6, 2) = 15

60 × 15 = 900

This proves the statement.

Answers: Verified (a) 360; (b) 900

Problem 16

In how many ways can the letters of the word 'COMPUTER' be arranged so that

a) all the vowels are always together?

b) the vowels may occupy only odd positions?

c) the relative positions of vowels and consonants are not changed?

Solution

COMPUTER has 8 distinct letters: C,O,M,P,U,T,E,R

Vowels: O, U, E (3 vowels)

Consonants: C, M, P, T, R (5 consonants)

Part (a)

Treat the 3 vowels as one block. Now we have 6 entities: (OUE), C, M, P, T, R

Arrange 6 entities: 6! = 720

Arrange 3 vowels within the block: 3! = 6

720 × 6 = 4320

Part (b)

Odd positions: 1,3,5,7 (4 positions)

Even positions: 2,4,6,8 (4 positions)

We need to place 3 vowels in 4 odd positions: C(4,3) = 4 ways to choose positions

Arrange 3 vowels in these positions: 3! = 6 ways

Arrange 5 consonants in remaining 5 positions: 5! = 120 ways

4 × 6 × 120 = 2880

Part (c)

The relative positions of vowels and consonants are not changed means:

Vowels must remain in positions where vowels originally were, and consonants where consonants were.

Original positions: C O M P U T E R (C,V,C,C,V,C,V,C)

So vowels must be in positions 2,5,7 and consonants in positions 1,3,4,6,8

Arrange 3 vowels in 3 vowel positions: 3! = 6

Arrange 5 consonants in 5 consonant positions: 5! = 120

6 × 120 = 720

Answers: (a) 4320; (b) 2880; (c) 720

Problem 17

Find the number of arrangements of the letters of the word "Laptop" so that

a) the vowels may never be separated.

b) all consonants may not be together.

c) they always begin with L and end with T.

d) they do not begin with L but always end with T.

Solution

Laptop has 6 letters: L, A, P, T, O, P

But P appears twice, so we have 6 letters with P repeated twice.

Vowels: A, O (2 vowels)

Consonants: L, P, T, P (4 consonants, with P repeated)

Part (a)

Treat the 2 vowels as one block. Now we have 5 entities: (AO), L, P, T, P

But P appears twice, so:

5! / 2! = 120 / 2 = 60

Arrange 2 vowels within the block: 2! = 2

60 × 2 = 120

Part (b)

Total arrangements of all letters: 6! / 2! = 720 / 2 = 360

Arrangements where all consonants are together:

Treat 4 consonants as one block: (L,P,T,P)

Now we have 3 entities: (consonant block), A, O

Arrange 3 entities: 3! = 6

Arrange 4 consonants within block: 4! / 2! = 24 / 2 = 12 (since P appears twice)

6 × 12 = 72

Arrangements where consonants are not together: Total - Together

360 - 72 = 288

Part (c)

Fix L at first position and T at last position.

Arrange remaining 4 letters (A,P,O,P) in middle 4 positions: 4! / 2! = 24 / 2 = 12

Answer: 12

Part (d)

Do not begin with L but end with T.

First position cannot be L, so we have 4 choices (A,P,O,P)

But P appears twice, so distinct choices: A, P, O (3 choices)

Fix T at last position.

Case 1: First letter is A or O (2 choices)

Then arrange remaining 4 letters (including both P's) in middle 4 positions: 4! / 2! = 12

Subtotal: 2 × 12 = 24

Case 2: First letter is P (1 choice)

Then arrange remaining 4 letters (A,O,P,T) but T is fixed at last, so actually arrange A,O,P in middle 3 positions: 3! = 6

Subtotal: 1 × 6 = 6

Total: 24 + 6 = 30

Answers: (a) 120; (b) 288; (c) 12; (d) 30

Problem 18

How many different words can be formed with all the letters of the word "Internet" if

a) each word is to begin with vowel?

b) each word is to end with consonant?

Solution

Internet has 8 letters: I, N, T, E, R, N, E, T

Frequency: I-1, N-2, T-2, E-2, R-1

Vowels: I, E, E (3 vowels, with E repeated)

Consonants: N, T, R, N, T (5 consonants, with N and T repeated)

Part (a)

Words beginning with vowel:

First letter must be a vowel: I or E

Case 1: First letter is I (1 choice)

Then arrange remaining 7 letters (N,T,E,R,N,E,T): 7! / (2! × 2! × 2!) = 5040 / 8 = 630

Case 2: First letter is E (1 choice, but E appears twice)

Then arrange remaining 7 letters (I,N,T,R,N,E,T): 7! / (2! × 2!) = 5040 / 4 = 1260

Total: 630 + 1260 = 1890

Part (b)

Words ending with consonant:

Last letter must be a consonant: N, T, or R

Case 1: Last letter is N (1 choice, but N appears twice)

Then arrange remaining 7 letters (I,N,T,E,R,E,T): 7! / (2! × 2! × 2!) = 5040 / 8 = 630

Case 2: Last letter is T (1 choice, but T appears twice)

Then arrange remaining 7 letters (I,N,T,E,R,N,E): 7! / (2! × 2! × 2!) = 5040 / 8 = 630

Case 3: Last letter is R (1 choice)

Then arrange remaining 7 letters (I,N,T,E,N,E,T): 7! / (2! × 2! × 2!) = 5040 / 8 = 630

Total: 630 + 630 + 630 = 1890

Answers: (a) 1890; (b) 1890

Answer Summary

Problem	Answer
1	60
2	720
3a	360, 180
3b	360, 60, 300
3c	1440, 9604
4a	5040
4b	144
4c	720
4d	576
5a	10080
5b	30240
6a	240
6b	480
7	27720
8a	840
8b	5040
8c	4989600
8d	34650
9	50
10a	5040
10b	144
11	1440
12	360
13a	1296
13b	450
13c	64
14	720, 600, 96
15	Verified: 360, 900
16a	4320
16b	2880
16c	720
17a	120
17b	288
17c	12
17d	30
18a	1890
18b	1890

↑

Class 12 Mathematics - Exercise 1.2 Solutions

Approved by Curriculum Development Centre (CDC), Nepal