Odisha State Board Elements of Mathematics Class 11 Solutions CHSE Odisha Chapter 8 Permutations and Combinations Ex 8(c) Textbook Exercise Questions and Answers.

## CHSE Odisha Class 11 Math Solutions Chapter 8 Permutations and Combinations Exercise 8(c)

Question 1.

Compute the following :

(i) ^{12}C_{3}

Solution:

^{12}C_{3} = \(\frac{(12) !}{3 ! 9 !}=\frac{12 \cdot 11 \cdot 10}{3 \cdot 2}\) = 220

(ii) ^{15}C_{12}

Solution:

^{15}C_{12} = \(\frac{(15) !}{(12) ! 3 !}=\frac{15 \cdot 14 \cdot 13}{3 \cdot 2}\)

= 5.7.13 = 455

(iii) ^{9}C_{4} + ^{9}C_{5}

Solution:

^{9}C_{4} + ^{9}C_{5} = \(\frac{9 !}{4 ! 5 !}+\frac{9 !}{5 ! 4 !}\)

\(=\frac{9 \cdot 8 \cdot 7 \cdot 6}{4 \cdot 3 \cdot 2 \cdot 1}\) × 2 = 252

(iv) ^{7}C_{3 } + ^{6}C_{4 } + ^{6}C_{3}

Solution:

^{7}C_{3 } + ^{6}C_{4 } + ^{6}C_{3} = ^{7}C_{3 } + ^{6}C_{4 } + ^{6}C4-1

= ^{7}C_{3} + ^{6+1}C_{4} = ^{7}C_{3} + ^{7}C_{4}

(∴ ^{n}c_{r} + ^{n}C_{r-1}– = ^{n+1}c_{r})

= ^{7}C_{4} + ^{7}C_{4-1} = ^{7+1}C_{4}

= ^{8}C_{4} = \(\frac{8 !}{4 !(8-4) !}=\frac{8 \cdot 7 \cdot 6 \cdot 5}{4 \cdot 3 \cdot 2 \cdot 1}\) = 70

(v) ^{8}C_{0} + ^{8}C_{1} + …….. + ^{8}c_{8}

Solution:

^{8}C_{0} + ^{8}C_{1} + …….. + ^{8}c_{8} = 2^{8} = 256

Question 2.

Solve :

(i) ^{n}C_{4} = ^{n}C_{11} ;

Solution:

^{n}C_{4} = ^{n}C_{11} ; (∴ n = 4 + 11 = 15)

(ii) ^{2n}C_{3} : ^{n}C_{3} = 44: 5

Solution:

^{2n}C_{3} : ^{n}C_{3} = \(\frac{44}{5}\)

⇒ \(\frac{2 n !}{2 n-3 !} / \frac{n !}{n-3 !}=\frac{44}{5}\)

Question 3.

Find n and r if ^{n}P_{r} = 1680, ^{n}C_{r} = 70.

Solution:

^{n}P_{r} = 1680, ^{n}C_{r} = 70

∴ \(\frac{{ }^n \mathrm{P}_r}{{ }^n \mathrm{C}_r}=\frac{1680}{70}\)

or, r ! = 24 = 4!

∴ r = 4

Again, ^{n}C_{r} = 70 or ^{n}C_{4} = 70

or, \(\frac{n !}{4 !(n-4) !}=70\)

or, n(n – 1) (n – 2) (n – 3)

= 70 × 4! = 7 × 10 × 4 × 3 × 2

= 8 × 7 × 6 × 5

or, n(n – 1) (n – 2) (n – 3)

or, 8(8 – 1) (8 – 2) (8 – 3)

∴ n = 8

Question 4.

How many diagonals can an n-gon(a polygon with n sides) have?

Solution:

A polygon of n – sides has n vertices.

∴ The number of st. lines joining the n-vertices is ^{n}C_{2}.

∴ The number of diagonals is ^{n}C_{2} – n

Question 5.

If a set A has n elements and another set B has m elements, what is the number of relations from A to B?

Solution:

If |A| = n, |B| = m

then |A × B| = mn

∴ The number of possible subsets of

A × B = 2^{mn}

∴ The number of relations from A to B is 2^{mn}.

Question 6.

From five consonants and four vowels, how many words consist of three consonants and two vowels?

Solution:

Words of consisting of 3 consonants and 2 vowels are to be formed from five consonants and 4 vowels.

∴ The number of ways = ^{5}C_{3} × ^{4}C_{2}

Again, 5 letters can be arranged among themselves in 5! ways.

∴ The total number of ways

= ^{5}C_{3} × ^{4}C_{2} × 5! = 10 × 6 × 120 = 7200.

Question 7.

In how many ways can a committee of four gentlemen and three ladies be formed out of seven gentlemen and six ladies?

Solution:

A committee of 4 gentlemen and 3 ladies is to be formed out of 7 gentlemen and 6 ladies.

∴ The number of ways in which the committee can be formed.

^{7}C_{4} × ^{6}C_{3} = \(\frac{7 \cdot 6 \cdot 5}{3.2} \times \frac{6 \cdot 5 \cdot 4}{3 \cdot 2}\) = 700

Question 8.

A bag contains 4 black and 5 white balls out of which 6 balls are drawn arbitrarily. In how many ways can this be done? Find also the number of ways such that at least 3 black balls can be drawn.

Solution:

A bag contains 4 black and 5 white balls out of which 6 balls are drawn arbitrarily.

∴ The number of ways in which balls are drawn \({ }^9 \mathrm{C}_6=\frac{9 \cdot 8 \cdot 7}{3 \cdot 2 \cdot 1}\) = 84 as the total number of balls is 9. If at least 3 black balls are drawn, then the drawing can be made as follows.

Black(4) | White(5) |

3 | 3 |

4 | 2 |

The number of ways in which at least 3 black balls are drawn

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

= (4 × 10) + (1 × 10) = 50

Question 9.

How many triangles can be drawn by joining the vertices of a decagon?

Solution:

A decagon has 10 vertices and 3 noncollinear points are required to be a triangle.

∴ The number of triangles formed by the joining of the vertices of a decagon is

^{10}C_{3} = \(\frac{10 !}{3 ! 7 !}=\frac{10 \cdot 9 \cdot 8}{3 \cdot 2 \cdot 1}\) = 120

Question 9.

How many triangles can be drawn by joining the vertices and the center of a regular hexagon?

Solution:

A regular hexagon has six vertices. Triangles are to be formed by joining the vertices and center of the hexagon. So there is a total of 7 points. So the number of triangles formed.

^{7}C_{3} = \(\frac{7 !}{3 ! 4 !}=\frac{7 \cdot 6 \cdot 5}{3 \cdot 2 \cdot 1}\) = 35

As the hexagon has 3 main diagonals, which pass through the center hence can not form 3 triangles.

∴ The required number of triangles 35 – 3 = 32

Question 11.

Sixty points lie on a plane, out of which no three points are collinear. How many straight lines can be formed by joining pairs of points?

Solution:

Sixty points lie on a plane, out of which no. 3 points are collinear. A straight line required two points. The number of straight lines formed by joining 60 points is

^{60}C_{2} = \(\frac{60 !}{2 \times 58 !}=\frac{60 \times 59}{2}\) = 1770

Question 12.

In how many ways can 10 boys and 10 girls sit in a row so that no two boys sit together?

Solution:

10 boys and 10 girls sit in a row so that no two boys sit together. So a boy is to be seated between two girls or at the two ends of the row. So the boys are to be sitted in 11 positions in ^{11}C_{10} ways. Again 10 boys and 10 girls can be arranged among themselves in 10! and 10! ways respectively.

∴ The total number of ways = ^{11}C_{10} × 10! × 10! = (11)! × (10)!

Question 13.

In how many ways can six men and seven girls sit in a row so that the girls always sit together?

Solution:

Six men and seven girls sit in a row so that the girls always sit together. Considering the 7 girls as one person, there are a total of 7 persons who can sit in 7! ways. Again the 7 girls can be arranged among themselves in 7! ways.

∴ The total number of arrangements

= 7! × 7!

= (7!)^{2}

Question 14.

How many factors does 1155 have that are divisible by 3?

Solution:

∴ In order to be a factor of 1155 divisible by 3, we have to choose one or two of 5, 7, and 11 along with 3 or 3 alone.

∴ The number of ways = ^{3}C_{1} + ^{3}C_{2} + ^{3}C_{0} = 2^{3} – 1 = 7

∴ The number of factors is 7 excluding 1155 itself.

Question 15.

How many factors does 210 have?

Solution:

∴ We can choose at least one 2, 3, 5, or 7 to be a factor of 210.

∴ The number of factors.

= ^{4}C_{1} + ^{4}C_{2} + ^{4}C_{3} + ^{4}C_{4} = 2^{4} – 1 = 15

∴ The number of factors is 210 is 15. (Including 215 itself and excluding 1).

Question 16.

If n is a product of k distinct primes what is the total number of factors of n?

Solution:

n is a product of k distinct primes.

∴ In order to be a factor, of n, we have chosen at least one of k distinct primes.

∴ The number of ways = ^{k}C_{1} + ^{k}C_{2} + ……… ^{k}C_{k-1} = 2^{k} – 1 – 1

∴ The number of factors of n is 2^{k} – 2.

(Excluding 1 as 1 is not prime. It is also not include n.)

Question 17.

If m has the prime factor decomposition P_{1}^{r1}, P_{2}^{r2} ….. P_{n}^{rn}, what is the total number of factors of m (excluding 1)?

Solution:

m has the prime factor decomposition P_{1}^{r1}, P_{2}^{r2} ….. P_{n}^{rn},

∴ m = P_{1}^{r1}, P_{2}^{r2} ….. P_{n}^{rn},

P_{1} is a factor of m which occurs r_{1} times. Each of the factors P_{1}^{r1} will give rise to (r_{1} + 1) factors.

Similarly

P_{2}^{r2} gives (r_{2} + 1) factors and so on.

∴ The total number of factors (r_{1} + 1) (r_{2} + 1) ….. (r_{n} + 1) – 1 (including m).

Question 18.

If 20! were multiplied out, how many consecutive zeros would it have on the right?

Solution:

If 20! were multiplied out, then the number of consecutive zeros on the right is 4. due to the presence of 4 x 5, 10, 14 x 15,20.

Question 19.

How many factors of 10,000 end with a 5 on the right?

Answer:

We have 1000 = 2^{4} × 5^{4} The factors of 10000 ending with 5 are 5, 5 × 5 = 25, 5 × 5 × 5 = 125

5 × 5 × 5 × 5 = 625

∴ There are 4 factors ending With 5.

Question 20.

A man has 6 friends. In how many ways can he invite two or more to a dinner party?

Solution:

A man has 6 friends. He can invite 2 or more of his friends to a dinner party.

∴ He can invite 2, 3, 4, 5, or 6 of his friends in

^{6}C_{2} + ^{6}C_{3} + ^{6}C_{4} + ^{6}C_{5} + ^{6}C_{6} = 2^{6} – ^{6}C_{0} – ^{6}C_{1}

= 64 – 7 = 57 ways.

Question 21.

In how many ways can a student choose 5 courses out of 9 if 2 courses are compulsory?

Solution:

A student is to choose 5 courses out of 9 in which 2 courses are compulsory. as 2 courses are compulsory, he is to choose 3 courses out of 7 courses in ^{7}C_{3} = 35 ways.

Question 22.

In how many ways can a student choose five courses out of the courses? C_{1}, C_{2}, …………. C_{9} if C_{1}, C_{2} are compulsory and C_{6}, C_{8} cannot be taken together?

Solution:

A student chooses five courses out of the courses? C_{1}, C_{2}, …………. C_{9} if C_{1}, C_{2} are compulsory and C_{6}, C_{8} cannot be taken together.

∴ He is to choose 3 courses out of C_{3}, C_{4}, ………… ,C_{8}, C_{9}.

Without taking any restrictions 3 courses out of C_{3}, C_{4}, …… C_{9}, i.e. from 7 courses in ^{7}C_{3} ways. If C_{6}, C_{8} are taken together then one course only to be choosen from C_{3}, C_{4}, C_{5}, C_{7}, C_{9} by ^{5}C_{1} ways. Hence required number of ways.

= ^{7}C_{3} – ^{5}C_{1}

= \(\frac{7 \times 6 \times 5}{3 \times 2 \times 1}\) – 5

= 35 – 5 = 30 ways.

Question 23.

A cricket team consisting of 11 players is to be chosen from 8 batsmen and 5 bowlers. In how many ways can the team be chosen so as to include at least 3 bowlers?

Solution:

A cricket team consisting of 1 1 player is to be chosen from 8 batsmen and 5 bowlers, as to include at least 3 bowlers.

The selection can be made as follows :

Batsmen(8) | Bowlers(5) |

8 | 3 |

7 | 4 |

6 | 5 |

The number of selections is

(^{8}C_{8} × ^{5}C_{3}) + (^{8}C_{7} × ^{5}C_{4}) + (^{8}C_{6} × ^{5}C_{5})

= (1 × 10) + (8 × 5) + (28 × 1)

= 10 + 40 + 28 = 78

Question 24.

There are n + r points on a plane out of which n points lie on a straight line L and out of the remaining r points that lie outside L, no three points are collinear. What is the number of straight lines that can be formed by joining pairs of their points?

Solution:

There are n + r points on a plane out of which n points are collinear and out of which r points are not collinear.

∴ We can form a straight line by joining any two points.

n-collinear points form one line and r-non-collinear points form ^{r}C_{2} lines.

Again, each of the r non-collinear points when joined to each of the noncollinear points, forms n lines.

∴ The number of such limes is r x n.

∴ The total number of lines

Question 25.

There are 10 books in a shelf with different titles; five of these have red covers and others have green covers. In how many ways can these be arranged so that the red books are placed together?

Solution:

There are 10 books in a shelf with different titles, 5 of these are red covers and others are green covers considering 5 red-covered books as one book, we have a total of 6 books which can be arranged in 6! ways. The five red cover books are arranged among themselves in 5! ways.

∴ The total number of arrangements

= 5! x 6!