**Chapter 16: Permutations**** – Exercise 16.2**

**Permutations – Exercise – 16.2 – Q.1**

Here the teacher is to perform two jobs.

(i) selecting a boy among 27 boys, and

(ii) selecting a girl among 14 girls.

The first of these can be performed in 27 ways and the second in 14 ways.Therefore by the fundamental principle of multiplication, the required number of ways is

27 × 14 = 378

Hence, the teacher can make the selection of a boy a girl in 378 ways.

**Permutations – Exercise – 16.2 – Q.2**

Here the person is to perform three jobs.

(i) selecting a ball pen from 12 ball pens

(ii) selecting a fountain pen from 10 fountain pens, and

(iii) selecting a pencil from 5 pencils.

The first of these can be performed in 12 ways, the second in 10 ways and the third in 5 ways.

Therefore by the fundamental principle of multiplication, the required number of ways is

12 × 10 × 5 = 600

Hence, the person can make the selection of a fountain pen, ball pen and pencil in 600 ways.

**Permutations – Exercise – 16.2 – Q.3**

From Bombay to Delhi there are three routs; air rail and road

Therefore by the fundamental principle of multiplication, the required number of ways are 2 × 3 = 6

Hence, total number of different kinds routes are 6.

**Permutations – Exercise – 16.2 – Q.4**

The mint has to perform two jobs,

(i) selecting the number of days in the february month (there can be 28 days or 29 days), and

(ii) selecting the first day of february.

The first job can be completed in 2 ways the second can be performed in 7 ways by selecting any one of the seven days of a week.

Thus, the required number of plates = 2 × 7 = 14

Hence, total number of calendars = 7 × 2 = 14

**Permutations – Exercise – 16.2 – Q.4(i)**

Total number of letters = 7

Total number of letter boxes = 4.

∴ Total number of ways in which 7 letters be posted in 4 letter boxes = 4 × 4 × 4 × 4 × 4 × 4 × 4 = 4^{7}

**Permutations – Exercise – 16.2 – Q.5**

Total number of parcels = 4

Total number of post-offices = 5

Since a parcel can be sent to any one of the five post offices.

So, the required number of ways = 5 × 5 × 5 × 5

= 5^{4}

= 625

Hence, total number of ways is 625.

**Permutations – Exercise – 16.2 – Q.6**

Since toss of each coin can result in 2 ways.

When coin is tossed five times, the total number of outcomes

= 2 × 2 × 2 × 2 × 2

= 32

Hence, required number of ways is 32

**Permutations – Exercise – 16.2 – Q.7**

The number of ways to examinee answer a true/false type question is 2.

∴ the number of ways for an examinee to answer a set of ten true/false type questions = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024.

Hence, the required number of ways is 1024.

**Permutations – Exercise – 16.2 – Q.8**

The total number of ways to make attempt to open the lock = 10 × 10 × 10 = 1000.

The number of ways to successfully open the lock = 1

∴ The number of ways to make an unsuccessful attempt to open the lock = 1000 - 1 = 999.

Hence, required number of ways to make an unsuccessfully attempt to the open the lock is 999.

**Permutations – Exercise – 16.2 – Q.9**

Each one of the first three questions can be answered in 4 ways.

∴ The total number of ways to answered the first

three question = 4 × 4 × 4 = 64

Each of the next three question can be answered in 2 ways.

∴ The total number of ways the answered the next three questions = 2 × 2 × 2 = 8

so, total number of sequences at answers = 64 × 8 = 512

**Permutations – Exercise – 16.2 – Q.10**

There are 5 books on mathematics and 6 books on physics in a book shop.

The number of ways to select a mathematics book = 5

The number of ways to select a physics book = 6

Now,

(i) Number of ways in which a student can buy a mathematics book and a physics book = 5 × 6 = 30

(ii) Number of ways in which a student buy either a mathematics book or a physics book = 5 + 6 = 11

**Permutations – Exercise – 16.2 – Q.11**

Since there are 7 flags of different colours, therefore, first flag can be selected in 7 ways.

Now, the second flag can be selected from any one of the remaining flags in 6 ways.

Hence, by the fundamental principle of multiplication, the number of flag is 7 × 6 = 42

**Permutations – Exercise – 16.2 – Q.12**

A boy can be selected from the first team in 6 ways, and from the second in 5 ways. so, number of single matches between the boys of two teams = 6 × 5 = 30.

similarly, the number of single matches between the girls of two teams = 4 × 3 = 12.

so, total number of matches = 30 + 12 = 42.

**Permutations – Exercise – 16.2 – Q.13**

Clearly, the total number of ways to select first three prizes is equal to the 3 students from 12 students.

Number of ways to select the three prizes = 12 × 11 × 10 = 1320

**Permutations – Exercise – 16.2 – Q.14**

There are 3 ways to choose the first form and corresponding to each such way there are 5 ways of selecting the common difference.

So, required number of A.P.'s

= 3 × 5

= 15

**Permutations – Exercise – 16.2 – Q.15**

Clearly the number of ways to appoint one principal, one vice-principal and the teacher- incharge is equal to the number of ways to select the three teachers from the 36 teachers.

∴ Number of ways to appointed 3 teachers = 36 × 35 × 34 = 42840

Hence, the number of ways to appoint one principal, one vice-prindpal and the teacher-incharge is equal to 42840.

**Permutations – Exercise – 16.2 – Q.16**

We have to form all possible 3-digit numbers with distinct digits.

we cannot have 0 at the hundred's place. so, the hundred's place can be filled with any of the 9 digits 1, 2 ,3, 4....,9.

so, there are 9 ways of filling the hundred's place.

Now, 9 digits are left including 0, so, ten's place can be filled with any of the remaining 9 digits in 9 ways. now, the unit's place can be filled which in any of the remaining 8 digits. so, there are 8 ways of filling the unit's place.

Hence, the total number of required numbers = 9 × 9 × 8 = 648

**Permutations – Exercise – 16.2 – Q.17**

We cannot have a 0 at the hundred's place. So, the hundred's place can be filled with any of the 9 digits

1, 2, 3, .............., 9

So, there are 9 ways of filling the hundred's place.

Ten's place can be filled with any 10 digits in 10 ways.

Now, the unit's place can be filled with any 10 digits in 10 ways.

Hence, the total number of required numbers = 9 × 10 × 10 = 900

**Permutations – Exercise – 16.2 – Q.18**

The three digit numbers ate 100 to 999 inclusive so there are

999 - 100 + 1 = 999 - 99 = 900

So, 900 three digit numbers

If half of all numbers is odd then half of 900 is 450, there am 450 odd positive 3 digit numbers

**Permutations – Exercise – 16.2 – Q.19(i)**

Zero cannot be first digit of the license plates.

This means the first digit can be selected from the 9 digits 1,2,3,4 .........., 9

So, there are 9 ways of filling the first digit of the license plates.

Now, 9 digits are left including 0. So, second place can be filled with any of the remaining 9 digits in 9 ways.

The third place of the license plates can be filled with in any of the remaining 8 digits. So, there are 8 ways of filling the third place.

The fourth place of the license plates can be filled with in any of the remaining 7 digits. So, there are 7 ways at filling the fourth place.

The last place of the license plates can be filled with in any of the remaining 6 digits. So, there are 6 ways of filling the fourth place.

Hence, the total number of ways = 9 × 9 × 8 × 7 × 6 = 27216