**Chapter 3 Permutations and Combination Ex 3.6**

## Chapter 3 Permutations and Combination Ex 3.6

**Question 1.Find the value of(a) **

^{15}C

_{4}Solution:

**(b) ^{80}C_{2}**

Solution:

**(c) ^{15}C_{4 }+^{15}C_{5}**

Solution:

**(d) ^{20}C_{16 }–^{19}C_{16}**

Solution:

**Question 2.Find n ifSolution:**

Solution:

Solution:

**Question 3.Solution:**

∴ 2r(2r – 1) (2r – 2) (2r – 3) = 14 × 12 × 10

∴ 2r(2r – 1) (2r – 2) (2r – 3) = 8 × 7 × 6 × 5

Comparing on both sides, we get

∴ r = 4

**Question 4.Find n and r if,Solution:**

Solution:

**Question 5.**

**Solution:**

**Question 6.Solution:**

**Question 7.Find the number of ways of drawing 9 balls from a bag that has 6 red balls, 8 green balls, and 7 blue balls so that 3 balls of every colour are drawn.Solution:**

9 balls are to be selected from 6 red, 8 green, 7 blue balls such that the selection consists of 3 balls of each colour.

∴ Number of ways selection can be done if the selection consists of 3 balls of each colour

**Question 8.Find the number of ways of selecting a team of 3 boys and 2 girls from 6 boys and 4 girls.Solution:**
There are 6 boys and 4 girls.

A team of 3 boys and 2 girls is to be selected.

∴ Number of ways the team can be selected

**Question 9.After a meeting, every participant shakes hands with every other participants. If the number of handshakes is 66, find the number of participants in the meeting.Solution:**

Let there be n participants present in the meeting.

A handshake occurs between 2 persons.

132 = n (n – 1)

n(n – 1) = 12 × 11

Comparing on both sides, we get n = 12

∴ 12 participants were present at the meeting.

**Question 10.If 20 points are marked on a circle, how many chords can be drawn?Solution:**

To draw a chord we need to join two points on the circle.

There are 20 points on a circle.

∴ Total number of chords possible from these points

**Question 11.Find the number of diagonals of an n-sided polygon. In particular, find the number of diagonals when(i) n = 10(ii) n = 15(iii) n = 12(iv) n = 8Solution:**
In n-sided polygon, there are ‘n’ points and ‘n’ sides.

∴ Through ‘n’ points we can draw

^{n}C

_{2}lines including sides.

∴ Number of diagonals in n sided polygon =

^{n}C

_{2}– n (n = number of sides)

**Question 12.There are 20 straight lines in a plane so that no two lines are parallel and no three lines are concurrent. Determine the number of points of intersection.Solution:**

There are 20 lines such that no two of them are parallel and no three of them are concurrent.

Since no two lines are parallel, they intersect at a point.

∴ Number of points of intersection if no two lines are parallel and no three lines are concurrent =

^{20}C

_{2}

= 190

**Question 13.Ten points are plotted on a plane. Find the number of straight lines obtained by joining these points if (a) no three points are collinear (b) four points are collinearSolution:**

There are 10 points on a plane.

(a) When no three of them are collinear.

A line is obtained by joining 2 points.

∴ Number of lines passing through these points =

^{10}C

_{2}

= 5 × 9

= 45

(b) When 4 of them are collinear.

If no three points are collinear, we get a total of ^{10}C_{2} = 45 lines by joining them. …..[From (i)]

Since 4 points are collinear, only one line passes through these points instead of ^{4}C_{2} lines.

∴ ^{4}C_{2} – 1 extra lines are included in 45 lines.

Number of lines passing through these points

= 45 – 6 + 1

= 40

**Question 14.Find the number of triangles formed by joining 12 points if(a) no three points are collinear(b) four points are collinearSolution:**

There are 12 points on the plane.

(a) When no three of them are collinear.

A triangle can be drawn by joining any three non-collinear points.

∴ Number of triangles that can be obtained from these points =

^{12}C

_{3}

= 220

(b) When 4 of these points are collinear.

If no three points are collinear, total we get ^{12}C_{3} = 220 triangles by joining them. ……[From (i)]

Since 4 points are collinear, no triangle can be formed by joining these four points.

= 220 – 4

= 216

**Question 15.A word has 8 consonants and 3 vowels. How many distinct words can be formed if 4 consonants and 2 vowels are chosen?Solution:**

There are 8 consonants and 3 vowels.

= 3 ways.

Now, to form a word, these 6 ietters (i.e., 4 consonants and 2 vowels) can be arranged in

^{6}P

_{6}= 6! ways.

∴ Total number of words that can be formed = 70 × 3 × 6!

= 70 × 3 × 720

= 151200

∴ 151200 words of 4 consonants and 2 vowels can be formed.

**Question 16.Find n if,**

Check:

Check:

**Question 17.Solution:**

**Question 18.Solution:**

**Question 19.Solution:**

**Question 20.Find the differences between the greatest values in the following:**

∴ Difference between the greatest values of ^{13}C_{r} and ^{8}C_{r} = ^{13}C_{r} – ^{8}C_{r}

= 1716 – 70

= 1646

∴ Difference between the greatest values of ^{15}C_{r} and ^{11}C_{r} = ^{15}C_{r} – ^{11}C_{r}

= 6435 – 462

= 5973

**Question 21.In how many ways can a boy invite his 5 friends to a party so that at least three join the party?Solution:**

Boy can invite = (3 or 4 or 5 friends)

Consider the following table:

∴ Number of ways a boy can invite his friends to a party so that three or more of join the party = 10 + 5 + 1 = 16

**Question 22.A group consists of 9 men and 6 women. A team of 6 is to be selected. How many of possible selections will have at least 3 women?Solution:**
There are 9 men and 6 women.

A team of 6 persons is to be formed such that it consist of at least 3 women.

Consider the following table:

∴ No. of ways this can be done = 1680 + 540 + 54 + 1 = 2275

∴ 2275 teams can be formed if team consists of at least 3 women.

**Question 23.A committee of 10 persons is to be formed from a group of 10 women and 8 men. How many possible committees will have at least 5 women? How many possible committees will have men in majority?Solution:**
(i) A committee of 10 persons is to be formed from 10 women and 8 men such that the committee contains at least 5 women.

Consider the following table:

∴ Number of committees with at least 5 women

= 14112 + 14700 + 6720 + 1260 + 81

= 36873

(ii) Number of committees with men in majority = Total number of committees – (Number of committees with women in majority + women and men equal in number)

= ^{18}C_{10} – 36873

= ^{18}C_{10} – 36873

= 43758 – 36873

= 6885

**Question 24.A question paper has two sections. Section I has 5 questions and section II has 6 questions. A student must answer at least two questions from each section among 6 questions he answers. How many different choices does the student have in choosing questions?Solution:**

There are 11 questions, out of which 5 questions are from section I and 6 questions are from section II.

The student has to select 6 questions taking at least 2 questions from each section.

Consider the following table:

∴ Number of choices = 150 + 200 + 75 = 425

∴ In 425 ways students can select 6 questions, taking at least 2 questions from each section.

**Question 25.There are 3 wicketkeepers and 5 bowlers among 22 cricket players. A team of 11 player is to be selected so that there is exactly one wicketkeeper and at least 4 bowlers in the team. How many different teams can be formed?Solution:**

There are 22 cricket players, of which 3 are wicketkeepers and 5 are bowlers.

A team of 11 players is to be chosen such that exactly one wicket keeper and at least 4 bowlers are to be included in the team.

Consider the following table:

∴ Number of ways a team of 11 players can be selected

= 45045 + 6006

= 51051

**Question 26.Five students are selected from 11. How many ways can these students be selected if(a) two specified students are selected?(b) two specified students are not selected?Solution:**

5 students are to be selected from 11 students.

(a) When 2 specified students are included,

then remaining 3 students can be selected from (11 – 2) = 9 students.

= 84

∴ Selection of students is done in 84 ways when 2 specified students are included.

(b) When 2 specified students are not included, then 5 students can be selected from the remaining (11 – 2) = 9 students.

= 126

∴ Selection of students is done in 126 ways when 2 specified students are not included.