Class 9 Advanced Maths Chapter 5 Combinatorics Ex 5.1 Solutions

Students often refer to NCERT Class 9 Advanced Maths Solutions Chapter 5 Combinatorics Ex 5.1 to verify their answers.

Advanced Maths Class 9 Exercise 5.1 Solutions

Class 9 Advanced Maths Ex 5.1 Solutions

Question 1.
A restaurant offers 4 starters, 5 main courses and 3 desserts. In how many ways can a 3-course meal be ordered?
Solution:
A starter can be selected in four ways. A main course can be selected in five ways. A dessert in three ways.
So, by the fundamental principle of multiplication, number of ways of selecting a three-course meal = 4 × 5 × 3 = 60

Question 2.
There are 5 doors to enter a hall and 3 different doors to exit. In how many ways can a person enter and exit the hall?
Solution:
The person can enter the hall in five ways. He can exit in three ways.
By the fundamental principle of multiplication, the required number of ways = 5 × 3 = 15

Question 3.
A bicycle lock has 3 dials, each with digits 0 to 9. How many different lock combinations are possible if a digit can be repeated?
Solution:
Each digit can be selected in 10 ways.
By the fundamental principle of multiplication, the required number of ways = 10 × 10 × 10 = 1000

Question 4.
How many numbers between 2000 and 3000 can be formed from the digits 2, 3, 4, 5, 6, 7 when repetition of digits is not allowed?
Solution:
Numbers between 2000 and 3000 must start with 2. Digits available: 2, 3, 4, 5, 6, 7 Repetition is not allowed.
Thousands place = 2, fixed.
Remaining 3 places can be filled from 5 remaining digits. So, required number of ways = 5 × 4 × 3 = 60

Question 5.
How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 6 without repetition? What if the repetition of digits is allowed?
Solution:
The digit in the unit’s place should be 2, 4 or 6. The units place can be filled in 3 ways. The tens place in 4 ways and the hundreds place in 3 ways.

By the fundamental principle of multiplication, the required number of even numbers without repetition of the digits = 3 × 4 × 3 = 36

If the repetition of digits is allowed

The digit in the unit’s place should be 2, 4 or 6. The unit’s place can be filled in 3 ways. The tens place in 5 ways and the hundreds place in 5 ways.

By the fundamental principle of multiplication, the required number of even numbers = 3 × 5 × 5 = 75.

Question 6.
How many numbers are there between 100 and 1000 such that 9 is in the units place? How many numbers will be there if 9 is at the tens place? How will this number change if 9 is at the hundreds place. Do you see a pattern? Can you describe this in your own language?
Solution:
9 in the units place Units digit is fixed as 9
Hundreds digit: cannot be 0 ⇒ 9 choices (1 – 9)
Tens digit: 10 choices (0 – 9)
Total = 9 × 10 = 90

9 in the tens place
Tens digit is fixed as 9
Hundreds digit: 9 choices (1 – 9)
Units digit: 10 choices (0 – 9)
Total = 9 × 10 = 90

9 in the hundreds place
Hundreds digit is fixed as 9
Tens digit: 10 choices (0 – 9)
Units digit: 10 choices (0 – 9)
Total = 10 × 10 = 100

Observation / Pattern
When 9 is fixed in units or tens place, first digit still has restriction (cannot be 0), giving 9 choices, so total = 9 × 10.

When 9 is fixed in the hundreds place, remaining two places are completelv free, giving 10 × 10.

General Result
If 9 is in any position other than the first: 9 × 10^n-2
If 9 is in the first position: 10^n-1.