# Class 11 RD Sharma Solutions – Chapter 17 Combinations- Exercise 17.2 | Set 1

**Question 1. From a group of 15 cricket players, a team of 11 players is to be chosen. In how many ways can this be done?**

**Solution:**

We are given,

Total number of players = 15

Number of players to be chosen = 11

So, number of ways =

^{15}C_{11}=

=

= 15 × 7 × 13

= 1365

Therefore, the total number of ways of choosing 11 players out of 15 is 1365.

**Question 2. How many different boat parties of 8, consisting of 5 boys and 3 girls, can be made from 25 boys and 10 girls?**

**Solution:**

We are given,

Total number of boys = 25

Total number of girls = 10

Boat party of 8 is to be made from 25 boys and 10 girls, by selecting 5 boys and 3 girls.

So, number of ways =

^{25}C_{5}×^{10}C_{3}=

= 53130 × 120

= 6375600

Therefore, the total number of different boat parties that can be made is 6375600.

**Question 3. In how many ways can a student choose 5 courses out of 9 courses if 2 courses are compulsory for every student?**

**Solution:**

We are given,

Total number of courses = 9

Number of courses a students can have = 5

Out of 9 courses, 2 courses are compulsory. So, a student can choose from 3 (i.e., 5−2) courses only.

So, number of ways =

^{7}C_{3}=

=

= 7 × 5

= 35

Therefore, the total number of ways in which a student can choose the subjects is 35.

**Question 4. In how many ways can a football team of 11 players be selected from 16 players? How many of these will**

**(i) include 2 particular players?**

**(ii) exclude 2 particular players?**

**Solution:**

Total number of players = 16

Number of players to be selected = 11

So, number of ways =

^{16}C_{11}=

=

= 4×7×13×12

= 4368

(i) include 2 particular players?

As two particular players have to be kept in the team every time, we have to choose 9 players (11−2) out of the

14 players (16−2).

So, number of ways =

^{14}C_{9}=

=

= 7×13×11×2

= 2002

(ii) exclude 2 particular players?

As 2 particular players are already removed, we have to select 11 players out of the remaining 14 players (16−2).

So, number of ways =

^{14}C_{9}=

=

= 14×13×2

= 364

Therefore, the required number of ways are 4368, 2002, 364 respectively.

**Question 5. There are 10 professors and 20 students out of whom a committee of 2 professors and 3 students is to be formed. Find the number of ways in which this can be done. Further, find in how many of these committees:**

**(i) a particular professor is included.**

**(ii) a particular student is included.**

**(iii) a particular student is excluded.**

**Solution:**

We are given,

Total number of professor = 10

Total number of students = 20

It is given that a committee has to be formed by choosing 2 professors from 10 and 3 students from 20.

So, number of ways =

^{10}C_{2}×^{20}C_{3}=

=

= 5×9×10×19×6

= 51300 ways

(i) a particular professor is included.

As one particular professor has to be selected in the committee every time, we have to choose 1 professor (2−1) out of the 9 professors (10−1). Number of ways for choosing students remain the same.

So, number of ways =

^{9}C_{1}×^{20}C_{3}=

=

= 9×10×19×6

= 10260 ways

(ii) a particular student is included.

As one particular student has to be selected in the committee every time, we have to choose 2 professors (3−1) out of the 19 professors (20−1). Number of ways for choosing professors remain the same.

So, number of ways =

^{10}C_{2}×^{19}C_{2}=

=

= 5×9×19×9

= 7695 ways

(iii) a particular student is excluded.

As one particular student has been removed from selection panel, we have to choose 3 students out of 19 students (20−1). Number of ways for choosing professors remain the same.

So, number of ways =

^{10}C_{2}×^{19}C_{3}=

=

= 5×9×19×3×17

= 43605 ways

Therefore, the required number of ways are 51300, 10260, 7695, 43605 respectively.

**Question 6. How many different products can be obtained by multiplying two or more of the numbers 3, 5, 7, 11 (without repetition)?**

**Solution:**

Total number of ways will be sum of number of ways of multiplying two numbers, three numbers and four numbers.

So, number of ways =

^{4}C_{2}+^{4}C_{3}+^{4}C_{4}=

=

= 6 + 4 + 1

= 11

Therefore, the total number of ways of product is 11 ways.

**Question 7. From a class of 12 boys and 10 girls, 10 students are to be chosen for the competition, at least including 4 boys and 4 girls. The 2 girls who won the prizes last year should be included. In how many ways can the selection be made?**

**Solution:**

We are given,

Total number of boys = 12

Total number of girls = 10

Total number of girls for the competition = 10 + 2 = 12

As two particular girls have to be included, total students that can be selected = 10−2 = 8

So, number of ways = (

^{12}C_{6}×^{8}C_{2}) + (^{12}C_{5}×^{8}C_{3}) + (^{12}C_{4 }×^{8}C_{4})=

=

= (924 × 28) + (792 × 56) + (495 × 70)

= 25872 + 44352 + 34650

= 104874

Therefore, the total number of ways in which the selection can be made is 104874.

**Question 8. How many different selections of 4 books can be made from 10 different books, if**

**(i) there is no restriction**

**(ii) two particular books are always selected**

**(iii) two particular books are never selected**

**Solution:**

Total number of books = 10

Number of books to be selected = 4

**(i) there is no restriction**

Number of ways = Choosing 4 books out of 10 books =

^{10}C_{4}=

=

= 10×3×7

= 210

Therefore, the number of ways of selecting books if there is no restriction is 210.

**(ii) two particular books are always selected**

As we have to select two particular books every time, we can select 2 books (4−2) out of the remaining 8 books (10−2).

So, number of ways =

^{8}C_{2}=

=

= 4×7

= 28

Therefore, the number of ways of selecting books if two particular books are always selected is 28.

**(iii) two particular books are never selected**

As two particular books have been removed, we have to choose 4 books out of the remaining 8 books (10−2).

So, number of ways =

^{8}C_{4}=

=

= 7×2×5

= 70

Therefore, the number of ways of selecting books if two particular books are never selected is 70.

**Question 9. From 4 officers and 8 jawans in how many ways can 6 be chosen**

**(i) to include exactly one officer**

**(ii) to include at least one officer?**

**Solution:**

Total number of officers = 4

Total number of jawans = 8

Total number of selections to be made = 6

**(i) to include exactly one officer**

Out of 6 selections, only 1 has to be an officer. So remaining 5 have to be jawans.

So, number of ways = (

^{4}C_{1}) × (^{8}C_{5})=

=

= 4×4×7×2

= 224

Therefore, the number of ways of selection if only one officer has to be included is 224.

**(ii) to include at least one officer?**

Out of 6 selections, at least 1 has to be an officer. So, we can choose from 1 to all 4 officers in our selections. And number of jawans would adjust according to that.

So, number of ways = (

^{4}C_{1}×^{8}C_{5}) + (^{4}C_{2}×^{8}C_{4}) + (^{4}C_{3}×^{8}C_{3}) + (^{4}C_{4}×^{8}C_{2})=

=

= (4 × 56) + (6 × 70) + (4 × 56) + (1 × 28)

= 224 + 420 + 224 + 28

= 896

Therefore, the number of ways of selection if at least one officer has to be included is 896.

**Question 10. A sports team of 11 students is to be constituted, choosing at least 5 from class XI and at least 5 from class XII. If there are 20 students in each of these classes, in how many ways can the teams be constituted?**

**Solution:**

Total number of students in XI = 20

Total number of students in XII = 20

Number of students to be selected in a team = 11

Now, at least 5 from class XI and 5 from class XII have to be selected.

So, number of ways = (

^{20}C_{6}×^{20}C_{5}) + (^{20}C_{5}×^{20}C_{6)}= 2 (

^{20}C_{6}×^{20}C_{5})= 2

=

= 2×38760×15504

= 1201870080

Therefore, the number of ways in which the teams can be constituted is 1201870080.

**Question 11. A student has to answer 10 questions, choosing at least 4 from each of part A and part B. If there are 6 questions in part A and 7 in part B, in how many ways can the student choose 10 questions?**

**Solution:**

Total number of questions = 10

Questions in part A = 6

Questions in part B = 7

Number of questions a student can choose = 10

Now a student can choose at least 4 from each of part A and part B and total number of questions that he can choose must not exceed.

So, number of ways = (

^{6}C_{4}×^{7}C_{6}) + (^{6}C_{5}×^{7}C_{5}) + (^{6}C_{6}×^{7}C_{4})=

=

= (15×7) + (6×21) + (1×35)

= 105 + 126 + 35

= 266

Therefore, the number of ways of answering 10 questions is 266.

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