Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 6 Permutations and Combinations Ex 6.6 Questions and Answers.

## Maharashtra State Board 11th Commerce Maths Solutions Chapter 6 Permutations and Combinations Ex 6.6

Question 1.

Find the value of

(i) ^{15}C_{4}

Solution:

(ii) ^{80}C_{2}

Solution:

(iii) ^{15}C_{4} + ^{15}C_{5}

Solution:

(iv) ^{20}C_{16} – ^{19}C_{16}

Solution:

Question 2.

Find n if

(i) ^{6}P_{2} = n ^{6}C_{2}

Solution:

(ii) ^{2n}C_{3} : ^{n}C_{2} = 52 : 3

Solution:

(iii) ^{n}C_{n-3} = 84

Solution:

^{n}C_{n-3} = 84

∴ \(\frac{n !}{(n-3) ![n-(n-3)] !}\) = 84

∴ \(\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3) !}{(\mathrm{n}-3) ! \times 3 !}\) = 84

∴ n(n – 1) (n – 2) = 84 × 6

∴ n(n – 1) (n – 2) = 9 × 8 × 7

Comparing on both sides, we get

∴ n = 9

Question 3.

Find r if ^{14}C_{2r} : ^{10}C_{2r-4} = 143 : 10

Solution:

∴ \(\frac{14 \times 13 \times 12 \times 11}{2 \mathrm{r}(2 \mathrm{r}-1) \times(2 \mathrm{r}-2)(2 \mathrm{r}-3)}=\frac{143}{10}\)

∴ 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.

(i) ^{n}P_{r} = 720 and ^{n}C_{n-r} = 120

Solution:

(ii) ^{n}C_{r-1} : ^{n}C_{r} : ^{n}C_{r+1} = 20 : 35 : 42

Solution:

Question 5.

If ^{n}P_{r} = 1814400 and ^{n}C_{r} = 45, find r.

Solution:

∴ r! = 40320

∴ r! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1

∴ r! = 8!

∴ r = 8

Question 6.

If ^{n}C_{r-1} = 6435, ^{n}C_{r} = 5005, ^{n}C_{r+1} = 3003, find ^{r}C_{5}.

Solution:

Question 7.

Find the number of ways of drawing 9 balls from a bag that has 6 red balls, 5 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, 5 green, 7 blue balls such that the selection consists of 3 balls of each colour.

∴ 3 red balls can be selected from 6 red balls in ^{6}C_{3} ways.

3 green balls can be selected from 5 green balls in ^{5}C_{3} ways.

3 blue balls can be selected from 7 blue balls in ^{7}C_{3} ways.

∴ 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.

∴ 3 boys can be selected from 6 boys in ^{6}C_{3} ways.

2 girls can be selected from 4 girls in ^{4}C_{2} ways.

∴ Number of ways the team can be selected = ^{6}C_{3} × ^{4}C_{2}

= \(\frac{6 !}{3 ! 3 !} \times \frac{4 !}{2 ! 2 !}\)

= \(\frac{6 \times 5 \times 4 \times 3 !}{3 \times 2 \times 1 \times 3 !} \times \frac{4 \times 3 \times 2 !}{2 \times 2 !}\)

= 20 × 6

= 120

∴ The team of 3 boys and 2 girls can be selected in 120 ways.

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.

∴ Number of handshakes = ^{n}C_{2}

Given 66 handshakes were exchanged.

∴ 66 = ^{n}C_{2}

∴ 66 = \(\frac{n !}{2 !(n-2) !}\)

∴ 66 × 2 = \(\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2) !}{(\mathrm{n}-2) !}\)

∴ 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 = ^{20}C_{2}

= \(\frac{20 !}{2 ! 18 !}\)

= \(\frac{20 \times 19 \times 18 !}{2 \times 1 \times 18 !}\)

= 190

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

Solution:

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}

= \(\frac{20 !}{2 ! 18 !}\)

= \(\frac{20 \times 19 \times 18 !}{2 \times 1 \times 18 !}\)

= 190

Question 13.

Ten points are plotted on a plane. Find the number of straight lines obtained by joining these points if

(i) no three points are collinear

(ii) four points are collinear

Solution:

There are 10 points on a plane.

(i) No three of them are collinear:

Since a line is obtained by joining 2 points,

number of lines passing through these points if no three points are collinear = ^{10}C_{2}

= \(\frac{10 !}{2 ! 8 !}\)

= \(\frac{10 \times 9 \times 8 !}{2 \times 1 \times 8 !}\)

= 5 × 9

= 45

(ii) When 4 of them arc collinear:

∴ Number of lines passing through these points if 4 points are collinear

= ^{10}C_{2} – ^{4}C_{2} + 1

= 45 – \(\frac{4 !}{2 ! 2 !}\) + 1

= 45 – \(\frac{4 \times 3 \times 2 !}{2 \times 2 !}\) + 1

= 45 – 6 + 1

= 40

Question 14.

Find the number of triangles formed by joining 12 points if

(i) no three points are collinear

(ii) four points are collinear

Solution:

There are 12 points on the plane

(i) When no three of them are collinear:

Since 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}

= \(\frac{12 !}{3 ! 9 !}\)

= \(\frac{12 \times 11 \times 10 \times 9 !}{3 \times 2 \times 1 \times 9 !}\)

= 220

(ii) When 4 of these points are collinear:

∴ Number of triangles that can be obtained from these points = ^{12}C_{3} – ^{4}C_{3}

= 220 – \(\frac{4 !}{3 ! \times 1 !}\)

= 220 – \(\frac{4 \times 3 !}{3 !}\)

= 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:

Out of 8 consonants, 4 can be selected in ^{8}C_{4}

= \(\frac{8 !}{4 ! 4 !}\)

= \(\frac{8 \times 7 \times 6 \times 5 \times 4 !}{4 \times 3 \times 2 \times 1 \times 4 !}\)

= 70 ways

From 3 vowels, 2 can be selected in ^{3}C_{2}

= \(\frac{3 !}{2 ! 1 !}\)

= \(\frac{3 \times 2 !}{2 !}\)

= 3 ways

Now, to form a word, these 6 letters (i.e., 4 consonants and 2 vowels) can be arranged in ^{6}P_{6} i.e., 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.