Class 9 Advanced Maths Chapter 5 Combinatorics Ex 5.1 Solutions

Accurate and updated NCERT Class 9 Advanced Maths Solutions Chapter 5 Combinatorics Ex 5.1 help students build a strong mathematical foundation.

Ex 5.1 Class 9 Advanced Maths Solutions

Advanced Maths Class 9 Exercise 5.1 Solutions

Exercise 5.1 Class 9 Advanced Maths 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:
There are 3 courses to be chosen, namely Starter, Main course and Dessert.
Since, we have to order a 3-course meal.
So, Starter can be filled in 4 ways, Main course can be filled in 5 ways and Dessert can be filled in 3 ways.

Starter	Main course	Dessert
↓	↓	↓
4 ways	5 ways	3 ways

Thus, by the fundamental principle of multiplication, the number of ways in which the 3-course meal can be ordered
= 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:
There are two actions to be performed, namely Entrance and Exit.
Since, a person has to enter and then exit the hall.
So, the Entrance can be filled in 5 ways and the Exit can be filled in 3 ways.

Entrance	Exit
↓	↓
5 ways	3 ways

Thus, by the fundamental principle of multiplication, the number of ways in which a person can enter and exit the hall = 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:
There are 10 digits available for each dial, namely 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9.

Since repetition of digits is allowed and we have to form a 3-dial combination.

So, the first dial can be filled in 10 ways, the second dial can be filled in 10 ways, and the third dial can also be filled in 10 ways.

First dial	Second dial	Third dial
↓	↓	↓
10 ways	10 ways	10 ways

Thus, by the fundamental principle of multiplication, the number of different lock combinations possible = 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:
There are 6 digits available namely 2, 3, 4, 5, 6 and 7.
We need to form a 4-digit number between 2000 and 3000.
For a number to be between 2000 and 3000, it must start with the digit 2.
Since, repetition of digits is not allowed.
The thousand’s place can be filled in only 1 way (using the digit 2). After using 2, we are left with 5 digits (3, 4, 5, 6, 7). So, the hundred’s place can be filled in 5 ways.
The ten’s place can be filled in 4 ways.
The unit’s place can be filled in 3 ways.

Thousand’s place	Hundred’s place	Ten’s place	Unit’s place
↓	↓	↓	↓
1 way (digit 2)	5 ways	4 ways	3 ways

Thus, by the fundamental principle of multiplication, the number of such 4-digit numbers that can be formed = 1 × 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:
There are 5 digits available namely 1, 2, 3, 4 and 6.
We need to form a 3-digit even number without repetition.
For a number to be even, the unit’s place must be occupied by an even digit. From the given set, the even digits are 2, 4 and 6.
When repetition of digits is not allowed
The unit’s place can be tilled in 3 ways (using 2, 4, or 6).
After filling the unit’s place, we are left with 4 digits. So, the Ten’s place can be filled in 4 ways.
The hundred’s place can be filled in 3 ways.

Hundred’s place	Ten’s place	Unit’s place
↓	↓	↓
3 ways	4 ways	3 ways

Thus, by the fundamental principle of multiplication, the number of 3-digit even numbers = 3 × 4 × 3 = 36

When repetition of digits is allowed
The unit’s place can still be filled in 3 ways (using 2, 4 or 6) to keep the number even.
Since, digits can be repeated, the ten’s place can be filled in all 5 ways.
Similarly, the hundred’s place can also be filled in all 5 ways.

Hundred’s place	Ten’s place	Unit’s place
↓	↓	↓
5 ways	5 ways	3 ways

According to the fundamental principle of multiplication, total number of ways = 5 × 5 × 3 = 75.
Hence, without repetition there are 36 numbers and with repetition allowed there are 75 numbers.

Question 6.
How many numbers are there between 100 and 1000 such that 9 is in the unit’s place? How many numbers will be there if 9 is at the ten’s place? How this number will change, if 9 is at the hundred’s place. Do you see a pattern? Can you describe this in your language?
Solution:
We have to find the count of 3-digit numbers between 100 and 1000 under different conditions using digits 0 to 9. Since, the question does not specify otherwise, repetition of digits is allowed.

Case I When 9 is in the unit’s place.
To form a 3-digit number, the hundred’s place can be filled by any digit from 1 to 9 (except 0) and the ten’s place can be filled by any digit from 0 to 9.
∴ Number of ways to fill the hundred’s place = 9,
Number of ways to fill the ten’s place = 10
and digit 9 is fixed at the unit’s place, so number of ways to fill it = 1.

Hundred’s place	Ten’s place	Unit’s place
↓	↓	↓
9 ways	10 ways	1 ways

∴ Total numbers with 9 at unit’s place = 9 × 10 × 1 = 90.
Case II When 9 is in the ten’s place.
Similarly, the hundred’s place can be filled in 9 ways (1 – 9) and the unit’s place can be filled in 10 ways (0 – 9).
∴ Total numbers with 9 at ten’s place = 9 × 1 × 10 = 90.
Case III When 9 is in the hundred’s place
Here, the hundred’s place is fixed with digit 9. The ten’s and unit’s places can each be filled in 10 ways (0 – 9).
∴ Total numbers with 9 at hundred’s place = 1 × 10 × 10 = 100.

Pattern and Description
We observe that for the unit’s and ten’s places, the count is the same (90), but for the hundred’s place, it increases to 100.
This happens because the hundred’s place cannot be 0, which restricts the choices for other cases. However, when 9 is fixed at the hundred’s place, that restriction is already satisfied, allowing all 10 digits for the remaining positions.

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