Skip to content
Related Articles

Related Articles

Improve Article

Class 11 RD Sharma Solutions – Chapter 17 Combinations- Exercise 17.2 | Set 3

  • Last Updated : 19 Jun, 2021

Question 23. We wish to select 6 persons from 8, but if the person A is chosen, then B must be chosen. In how many ways can the selection be made?

Solution:

Total number of persons = 8

Number of persons to be selected = 6

It is given that if the person A is chosen, then B must be chosen. Therefore, A and B are chosen together.

So, number of selections = 6C4



=\frac{6!}{4!2!}

=\frac{6×5}{2×1}

= 15

Also the number of selections in which A and B are not chosen = 7C6

=\frac{7!}{6!1!}

= 7

Therefore, total number of ways = 15 + 7 = 22 ways.

Question 24. In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?

Solution:



Total number of boys = 5

Total number of boys = 4

Number of boys to be selected = 3

Number of girls to be selected = 3

Hence, we have to choose 3 girls from 4 girls and 3 boys from 5 boys.

So, number of ways = 5C3 × 4C3

=\frac{5!}{3!2!}×\frac{4!}{3!1!}

=\frac{5×4}{2×1}×\frac{4}{1}

= 40

Therefore the required number of ways is 40.



Question 25. Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour.

Solution:

We are given, 6 red balls, 5 white balls and 5 blue balls. We have to choose 3 balls of each colour.

Therefore, we can select 3 red balls from 6 red balls, 3 white balls from 5 white balls and 3 blue balls from 5 blue balls.

So, number of ways = 6C3 × 5C3 × 5C3

=\frac{6!}{3!3!}×\frac{5!}{3!2!}×\frac{5!}{3!2!}

=\frac{6×5×4}{3×2×1}×\frac{5×4}{2×1}×\frac{5×4}{2×1}

= 20×10×10

= 2000

Therefore, the required number of ways is 2000.

Question 26. Determine the number of 5 cards combinations out of a deck of 52 cards if there is exactly one ace in each combination.

Solution:



Out of a deck of 52 cards, we have to choose 5 cards combinations where we want exactly one ace in each combination.

We know there are 4 aces in a deck of 52 cards. So, we have to choose 1 ace from 4 aces and selection of remaining cards will be from remaining 48 cards.

So, number of ways = (4C1 × 48C4)

=\frac{4!}{1!3!}×\frac{48!}{4!44!}

=\frac{4}{1}×\frac{48×47×46×45}{4×3×2×1}

= 778320

Therefore, number of ways of selecting 5 card combinations if there is exactly one ace in each combination is 778320.

Question 27. In how many ways can one select a cricket team of eleven from 17 players in which only 5 persons can bowl if each cricket team of 11 must include exactly 4 bowlers?

Solution:

Total number of players = 17

Total number of bowlers = 5



Number of bowlers to be selected = 4

As number of bowlers are 5, therefore number of batsmen = 17−5 = 12

To make a team of 11 players, we have to choose 7 batsmen from 12 batsmen and 4 bowlers from 5 bowlers.

So, number of ways = 12C7 × 5C4

=\frac{12!}{7!5!}×\frac{5!}{4!1!}

=\frac{12×11×10×9×8}{5×4×3×2×1}×\frac{5}{1}

= 3960

Therefore, number of ways of selecting the team is 3960.

Question 28. A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.

Solution:

Total number of black balls = 5

Total number of red balls = 6

Now we have to choose 2 black balls from 5 black balls and 3 red balls from 6 red balls.

So, number of ways = 5C2 × 6C3

=\frac{5!}{2!3!}×\frac{6!}{3!3!}

=\frac{5×4}{2×1}×\frac{6×5×4}{3×2×1}

= 10×20

= 200

Therefore the required number of ways is 200.

Question 29. In how many ways can a student choose a program of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?

Solution:

Total number of courses = 9



Number of courses to be selected = 5

As 2 courses have been made compulsory, a student can only choose 3 courses (5−2) from remaining 7 courses (9−2).

So, number of ways = 7C3

=\frac{7!}{3!4!}

=\frac{7×6×5}{3×2×1}

= 35

Therefore, the required number of ways is 35.

Question 30. A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of

(i) exactly 3 girls?

(ii) at least 3 girls?

(iii) at most 3 girls?

Solution:

Total number of boys = 9

Total number of girls = 4

Number of persons to be selected = 7

(i) exactly 3 girls

If we want exactly 3 girls in the committee, we have to choose 3 girls from 4 girls and 4 boys from 9 boys.

So, number of ways = 9C4 × 4C3

=\frac{9!}{4!5!}×\frac{4!}{3!1!}

=\frac{9×8×7×6}{4×3×2×1}×\frac{4}{1}

= 9×8×7



= 504

Therefore, the number of ways of selecting a committee in which exactly 3 girls are required is 504.

(ii) at least 3 girls?

As the required number of girls can be 3 or more than 3, we can choose 3 or 4 girls and adjust the number of boys according to it.

So, number of ways = (9C4 × 4C3) + (9C3 × 4C4)

=(\frac{9!}{4!5!}×\frac{4!}{3!1!})+(\frac{9!}{3!6!}×\frac{4!}{4!0!})

=(\frac{9×8×7×6}{4×3×2×1}×\frac{4}{1})+(\frac{9×8×7}{3×2×1}×\frac{4}{1})

= 504 + 84

= 588

Therefore, the number of ways of selecting a committee in which at least 3 girls are required is 588.

(iii) at most 3 girls.

As the required number of girls can be 3 or less than 3, we can choose 3 or 2 or 1 or 0 number of girls and adjust the number of boys according to it.

So, number of ways = (4C0 × 9C7) + (4C1 × 9C6) + (4C2 × 9C5) + (4C3 × 9C4)

=(\frac{4!}{4!0!}×\frac{9!}{7!2!})+(\frac{4!}{1!3!}×\frac{9!}{6!3!})+(\frac{4!}{2!2!}×\frac{9!}{5!4!})+(\frac{4!}{3!1!}×\frac{9!}{4!5!})

=(\frac{1}{1}×\frac{9×8}{2×1})+(\frac{4}{1}×\frac{9×8×7}{3×2×1})+(\frac{4×3}{2×1}×\frac{9×8×7×6}{4×3×2×1})+(\frac{4}{1}×\frac{9×8×7×6}{4×3×2×1})

= 36 + 336 + 756+ 504

= 1632

Therefore, the number of ways of selecting a committee in which at most 3 girls are required is 1632.

Question 31. In an examination, a question paper consists of 12 questions divided into two parts Part I and Part II, containing 5 and 7 questions respectively. A student is required to attempt 8 questions in all, selecting at least 3 from each part. In how many ways can a student select the questions?

Solution:

Number of questions in Part I = 5



Number of questions in Part II = 7

Number of questions to be attempted = 8

Now a student has to attempt at least 3 questions from each part and 8 in total.

So, number of ways = (5C3 × 7C5) + (5C4 × 7C4) + (5C5 × 7C3)

=(\frac{5!}{3!2!}×\frac{7!}{5!2!})+(\frac{5!}{4!1!}×\frac{7!}{4!3!})+(\frac{5!}{5!0!}×\frac{7!}{3!4!})

=(\frac{5×4}{2×1}×\frac{7×6}{2×1})+(\frac{5}{1}×\frac{7×6×5}{3×2×1})+(\frac{1}{1}×\frac{7×6×5}{3×2×1})

= 210 + 175 + 35

= 420

Therefore, the number of ways a student can select the questions is 420.

Question 32. A parallelogram is cut by two sets of m lines parallel to its sides. Find the number of parallelogram thus formed.

Solution:

A parallelogram has 2 sets of parallel lines. Now m lines are parallel to these sets. Therefore, total number of lines for one set is (m+2) lines.

Therefore the total number of parallelograms formed = m+2C2 × m+2C2

=\frac{(m+2)!}{m!2!}×\frac{(m+2)!}{m!2!}

= (m+2)2 (m+1)2/4

Therefore, the number of parallelogram formed is (m+2)2 (m+1)2/4.

Question 33. Out of 18 points in a plane, no three are in the same straight line except five points which are collinear. How many

(i) straight lines can be formed by joining them?

Solution:

Given there are 18 points in a plane where 5 points are collinear.

Now we know, number of lines formed will be the difference between total number of lines formed by all 18 points and number of lines formed by collinear points added with 1.

Here, we add 1 because only 1 line can be formed by the given 5 collinear points.



So, number of ways = 18C25C2 + 1

=\frac{18!}{16!2!}-\frac{5!}{2!3!}+1

=\frac{18×17}{2×1}-\frac{5×4}{2×1}+1

= 144

Therefore number of straight lines formed by joining 18 points is 144.

(ii) triangles can be formed by joining them?

Solution:

Now three points are required to form a triangle.

Number of points that can be used to make triangles = 18−5 = 13

So, number of triangles = 13C3

=\frac{13!}{3!10!}

=\frac{13×12×11}{3×2×1}

= 13×22

= 286

Therefore number of triangles formed by joining 18 points is 286.

Attention reader! Don’t stop learning now. Participate in the Scholorship Test for First-Step-to-DSA Course for Class 9 to 12 students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :