Complete Exercise 1.1 Class 12 Solutions: Permutation & Combination.

Class 12 Mathematics - Exercise 1.1 Solutions

Permutations and Combinations

Table of Contents

• Question 1: Stadium Entrance and Exit
• Question 2: Hostel Doors
• Question 3: College Selection
• Question 4: City Travel with Different Roads
• Question 5: Multi-City Travel
• Question 6: Three-Digit Numbers
• Question 7: Three-Digit Numbers Less Than 500
• Question 8: Even Numbers with All Digits
• Question 9: Numbers Between 4000 and 5000
• Question 10: Three-Digit Numbers Divisible by 5
• Answer Summary

Question 1: Stadium Entrance and Exit

1. A football stadium has four entrance gates and nine exits. In how many different ways can a man enter and leave the stadium?

Using the Fundamental Principle of Counting:

Number of ways to enter = 4

Number of ways to leave = 9

Total ways = 4 × 9 = 36

So, a man can enter and leave the stadium in 36 different ways.

Question 2: Hostel Doors

2. There are six doors in a hostel. In how many ways can a student enter the hostel and leave by a different door?

Using the Fundamental Principle of Counting:

Number of ways to enter = 6

Number of ways to leave = 5 (since the student must use a different door)

Total ways = 6 × 5 = 30

So, a student can enter and leave by different doors in 30 ways.

Question 3: College Selection

3. In how many ways can a man send three of his children to seven different colleges of a certain town?

This is a permutation problem since the colleges are different:

Number of colleges = 7

Number of children to send = 3

Number of ways = P(7,3) = 7 × 6 × 5 = 210

So, there are 210 ways to send three children to seven different colleges.

Question 4: City Travel with Different Roads

4. Suppose there are five main roads between the cities A and B. In how many ways can a man go from a city to the other and return by a different road?

Using the Fundamental Principle of Counting:

Number of ways to go from A to B = 5

Number of ways to return from B to A = 4 (since a different road must be used)

Total ways = 5 × 4 = 20

So, there are 20 ways to go from A to B and return by a different road.

Question 5: Multi-City Travel

5. There are five main roads between the cities A and B and four between B and C. In how many ways can a person drive from A to C and return without driving on the same road twice?

Using the Fundamental Principle of Counting:

From A to B: 5 roads

From B to C: 4 roads

From C to B: 3 roads (since one road already used)

From B to A: 4 roads (since one road already used)

Total ways = 5 × 4 × 3 × 4 = 240

So, there are 240 ways to drive from A to C and return without using the same road twice.

Question 6: Three-Digit Numbers

6. How many numbers of at least three different digit numbers can be formed from the integers 1, 2, 3, 4, 5, 6?

We need to find 3-digit, 4-digit, 5-digit, and 6-digit numbers with all different digits:

3-digit numbers: P(6,3) = 6 × 5 × 4 = 120

4-digit numbers: P(6,4) = 6 × 5 × 4 × 3 = 360

5-digit numbers: P(6,5) = 6 × 5 × 4 × 3 × 2 = 720

6-digit numbers: P(6,6) = 6! = 720

Total = 120 + 360 + 720 + 720 = 1920

So, 1920 numbers with at least three different digits can be formed.

Question 7: Three-Digit Numbers Less Than 500

7. How many three different digit numbers less than 500 can be formed from the integers 1, 2, 3, 4, 5, 6?

For a number to be less than 500, the hundred's digit must be 1, 2, 3, or 4:

Case 1: Hundred's digit is 1, 2, 3, or 4 (4 choices)

Remaining digits: 5 choices for tens place, 4 choices for units place

Total = 4 × 5 × 4 = 80

So, 80 three-digit numbers less than 500 can be formed.

Question 8: Even Numbers with All Digits

8. Of the numbers formed by using all the figures 1, 2, 3, 4, 5 only once, how many are even?

For a number to be even, the units digit must be even (2 or 4):

Case 1: Units digit is 2

Remaining digits: 1, 3, 4, 5 can be arranged in 4! = 24 ways

Case 2: Units digit is 4

Remaining digits: 1, 2, 3, 5 can be arranged in 4! = 24 ways

Total even numbers = 24 + 24 = 48

So, 48 even numbers can be formed using all digits 1, 2, 3, 4, 5 exactly once.

Question 9: Numbers Between 4000 and 5000

9. How many different digit numbers between 4000 and 5000 can be formed with the digits 2, 3, 4, 5, 6, 7?

For a number to be between 4000 and 5000, it must be a 4-digit number starting with 4:

Thousand's digit must be 4 (1 choice)

Remaining digits: 2, 3, 5, 6, 7 (5 choices)

We need to choose 3 digits from these 5 and arrange them

Number of ways = P(5,3) = 5 × 4 × 3 = 60

So, 60 numbers between 4000 and 5000 can be formed.

Question 10: Three-Digit Numbers Divisible by 5

10. How many numbers of three different digit numbers can be formed from the integers 2, 3, 4, 5, 6? How many of them will be divisible by 5?

Part 1: Total three-digit numbers with different digits:

Total numbers = P(5,3) = 5 × 4 × 3 = 60

Part 2: Numbers divisible by 5 (units digit must be 5):

Units digit fixed as 5 (1 choice)

Hundred's digit: can be 2, 3, 4, 6 (4 choices)

Tens digit: remaining 3 choices

Total divisible by 5 = 4 × 3 = 12

So, 60 three-digit numbers can be formed, and 12 of them are divisible by 5.

Answer Summary

Question 1

36 ways

Question 2

30 ways

Question 3

210 ways

Question 4

20 ways

Question 5

240 ways

Question 6

1920 numbers

Question 7

80 numbers

Question 8

48 numbers

Question 9

60 numbers

Question 10

60 numbers, 12 divisible by 5

Class 12 Mathematics - Permutations and Combinations

Complete Exercise 1.1 Solutions