## Chapter 1: Permutation and Combination.

### Exercise: 1.1

Complete Exercise of Permutation and Combination - Exercise 1.1 : Class 12 Mathematics 2080 NEB.

**Read:**NEB Class 12 Mathematics All Chapter Exercise

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

Soln:

A man can enter the stadium in 4 ways. Again the man can leave the stadium in 9 ways.

So, total no.of ways with which a man enters and then leaves the stadium = 4* 9 = 36ways.

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

Soln:

There are 6 choices for a student to enter the hostel. There are 5 choices for a student to leave the hostel as different door is to be used. So, total no.of ways = 65 = 30.

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

Soln:

There are 7 choices for 1st son, 6 choices for 2nd son and 5 choices for 3rd son.

Now, by the basic principle of counting, the total number of ways of choice = 7*6*5 = 210.

**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?**

Soln:

A man can go from city A to city B in 5 ways. As he has to return by a different road, so he can return from city B to city A in 4 ways.

So, total no.of ways by which a man can go from city A to city B and returns by a different road = 5 * 4 = 20 ways.

**5. There are five main roads between the cities A and B and 4 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?**

Soln:

A person can go from city A to city B in 5 ways. Again, he can go from city B to city C in 4 ways. So, a person can go from city A to city C in 5* 4 = 20ways. The person has to return from C to A without driving on the same road twice, So, he can return from city C to city B in 3 ways and from city B to city A in 4 ways.

So, he can return from city C to city A in 3 * 4 = 12 ways.

So, Total no.of ways by which a person can go from city A to city C and return from city C to city A = 20 12 = 240 ways.

**6. How many numbers of three digits less than 500 can be formed from the integers 1, 2, 3, 4, 5, 6?**

Soln:

Numbers formed should be of at least 3 digits means they may be of 3 digits, 4 digits, 5 digits or 6 digits.

There are 6 choices for digit in the units place. There are 5 and 4 choices for digits in ten and hundred's place respectively.

So, total number of ways by which 3 digits numbers can be formed = 6.5.4 = 120

Similarly, the total no.of ways by which 4 digits

numbers can be formed = 6.5.4.3

360.

the total no. of ways by which 5 digits numbers

can be formed = 6.5.4.3.2 = 720.

The total no.of ways by which 4 digits numbers can be formed 6.5.4.3.2.1 = 720.

= So, total no.of ways by which the numbers of at least 3 digits can be formed = 120 + 360 + 720 + 720 = 1920.

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

Soln:

The numbers formed must be of three digits and less than 500, so the digit in the hundred's place should be 1,2,3 or 4. So, there are 4 choices for the digit in the hundred's place. There are 5 choice for the digit in the ten's place. There are 4 choices for the digit in the unit's place. So, no of ways by which 3 digits numbers les than 500 can be formed = 4.5.4 = 80.

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

Soln:

The numbers formed should be even. So, the digit in the unit's place must be 2 or 4. So, the digit in unit's place must be 2 or 4. So, for the digit in unit's place, there are 2 choices. So, after fixing the digit in the unit's place, remaining 4 figures can be arranged in P(4,4) ways.

`\frac{(4)!}{(4-4)!} = \frac{4!}{0!} = \frac{4*3*2*1}{1} = 24 ways`

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

Soln:

The numbers formed must be of 4 digits. The

digit in the thousand's place must always be 4. For this, there is only one choice. After that, n = 6 − 1 = 5, r = 4 − 1 = 3. Then remaining 5 figures can be placed in remaining 3 places in:

Or, P(5,3) ways =

`\frac{(5)!}{(5-3)!} = \frac{5!}{2!} = \frac{5*4*3*2*1}{2*1} = 60 ways`

So, Total no.of ways by which 4 digits numbers between 4,000 and 5,000 can be formed = 1 * 60

= 60.

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

Soln:

For the three digits numbers, there are 5 ways to

2nd

fill in the 1st place, there are 4 ways to fill in the place and there are 3 ways to fill in the 3rd place. By the basic principle of counting, number of three digits numbers = 5* 4 * 3 = 60.

Again, for three digit numbers which are divisible by 5, the number in the unit place must be 5. So, the unit place can be filled up in 1 way. After filling up the unit place 4 numbers are left. Ten's place can be filled up in 4 ways and hundredths place can be filled up in 3 ways. Then by the basic principle of counting, no.of 3 digits numbers which are divisible by 51*4*3 = 12.

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