Previous exercise – **Permutation and Combination – Exercise 4.6**

#### Permutation and Combination – Exercise 4.7

- Out of 7 consonants and 4 towels, how many words of 3 consonants and 2 vowels can be formed?
- In how many ways can 5 sportsmen can be selected from a group of 10?
- In how many ways a cricket team of eleven be selected from 17 players in which 5 players are bowlers and the cricket team must include 2 bowlers?
- How many (i) lines (ii) triangles can be drawn through 8 points on a circle?
- How many diagonals can be drawn in a (i) decagon (ii) icosagon
- A polygon has 44 diagonals. Find the number of sides
- If there are 6 periods in each working day of school, in how many ways can this be done when

(i) at least ladies are included?

(ii) at most 2 ladies are included?

- A committee of 5 is to be formed out of 6 men and 4 ladies are included?
- A sports team of 11 students is to be constituted choosing at least 6 from class IX and at least 5 from class X. IF there are 8 students in each of these classes, in how many ways can the team be constituted?
- From a group of 12 students, 8 are to be chosen for an excursion. There are 3 students who decide that either of them will join or none of them will join. In how many ways can the 8 be chosen?

#### Permutation and Combination – Exercise 4.7 – Solutions:

**Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?**

Solution:

Number of ways of selecting 3 consonants and 2 vowels out of 7 consonants and 4 vowels = (^{7}C_{3} x ^{4}C_{2}) = 210

Since each group has 5 letters – 3consonants and 2 vowels then, number of ways of arranging 5 letters among themselves = 5! = 1 x 2 x 3 x 4 x 5 = 120

Required number of ways = (Number of ways of selecting 3 consonants and 2 vowels out of 7 consonants and 4 vowels) x (number of ways of arranging 5 letters among themselves)

= 210 x 120 = 25200

**In how many ways can 5 sportsmen can be selected from a group of 10?**

Solution:

Here n= 10 and r = 5

Then, ^{n}C_{r} = ^{n!}/_{(n-r)!r!} = ^{10!}/_{(10-5)!5!} = ^{5!x6x7x8x9x10}/_{5! x 1x2x3x4x5} = ^{6x7x8x9x10}/_{1x2x3x4x5} = 252

Therefore, in 252 ways 5 sportsman can be selected from a group of 10.

**In how many ways a cricket team of eleven be selected from 17 players in which 5 players are bowlers and the cricket team must include 2 bowlers?**

Solution:

Given, a cricket team of eleven be selected from 17 players in which 5 players are bowlers and the cricket team must include 2 bowlers. Therefore, out of 17 players 5 are bowlers, then,

number of ways of selecting 11 cricketers out of 17 and 2 bowlers out of 5 = ^{12}C_{9} x ^{5}C_{2} = 2200

**How many (i) lines (ii) triangles can be drawn through 8 points on a circle?**

Solution:

(i) A line has 2 points. Then, the number of lines can be drawn through 8 points on a circle = ^{8}C_{2} = ^{8!}/_{(8-2)!2!} = ^{8!}/_{6!2!} = ^{7×8}/_{2} = 7 x 4 = 28

(ii) Triangle has 3 points. So the number of triangles can be drawn out of 8 points on a circle = ^{8}C_{3} = ^{8!}/_{(8-3)!3!} = ^{8!}/_{5!3!} = ^{5!x6x7x8}/_{5!x2x3} = ^{6x7x8}/_{2×3} = 56

**How many diagonals can be drawn in a (i) decagon (ii) icosagon**

Solution:

(i)A decagon has 10 vertices and 10 sides. A diagonal needs 2 points. Number of diagonals = number of straight lines formed – number of sides of polygon = ^{n}C_{r} – n .

Here, n = 10 and r = 2

Thus, ^{n}C_{r} – n = ^{n!}/_{(n-r)!r!} – n = ^{10!}/_{(10-2)!2!} – 10 = ^{8!x9x10}/_{8!x1x2} – 10 = 45 – 10 = 35

(ii) A icosagon has 20 vertices and 20 sides. A diagonal needs 2 points. Number of diagonals = number of straight lines formed – number of sides of polygon = ^{n}C_{r} – n .

Here, n = 20 and r = 2

Thus, ^{n}C_{r} – n = ^{n!}/_{(n-r)!r!} – n = ^{20!}/_{(20-2)!2!} – 10 = ^{18!x19x20}/_{18!x1x2} – 10 = 190 – 20 = 170

**A polygon has 44 diagonals. Find the number of sides**

Solution:

Given a polygon has 44 diagonals; we know a diagonal has 2 points. Number of diagonals = number of straight lines formed – number of sides of polygon

⇒ 44 = ^{n}C_{r} – n = ^{n!}/_{(n-r)!r!} – n

⇒ 44 = ^{n!}/_{(n-2)!2!} – n

⇒ 44×2! = ^{(n-2)!(n-1)n}/_{(n-2)!} – n

⇒ 88 = (n – 1)n – n

⇒ 88 = n^{2} – n – n

⇒88 = n^{2} – 2n

⇒n^{2} – 2n – 88 = 0

⇒n^{2} – 11n + 8n – 88 = 0

⇒n(n – 11) +8(n – 11) = 0

⇒ (n – 11)(n + 8) = 0

⇒ n = 11 or n = -8

Therefore, a polygon with 11 diagonals has 11 sides.

**If there are 6 periods in each working day of school, in how many ways can one arrange 6 subjects such that each subject is allowed at least one period?**

Solution:

Number of periods a school has is 6.

Therefore, in number of ways 6 periods can be arranged in such a way that 6 subjects get at least one period = 6! = 1 x 2 x 3 x 4 x 5 x 6 = 720.

**A committee of 5 is to be formed out of 6 men and 4 ladies are included. In how many ways this be done when**

**(i) at least 2 ladies are included?**

**(ii) at most 2 ladies are included?**

Solution:

A committee of 5 is to be formed out of 6 men and 4 ladies are included

(i)When at least 2 ladies are included that means committee may consists more than 2 women. i.e., committee may consist 3 women and 2 men such as ^{4}C_{3} x ^{6}C_{2} or committee may include 4 women and 1 man such as ^{4}C_{4} x ^{6}C_{1} or committee may include two women and 3 men ^{4}C_{2} x ^{6}C_{3}

(^{4}C_{3} x ^{6}C_{2}) + (^{4}C_{4} x ^{6}C_{1}) + (^{4}C_{2} x ^{6}C_{3}) = 186.

(ii) When at most 2 ladies are included i.e., committee may consist 2 women and 3 men such as ^{4}C_{2} x ^{6}C_{3} or committee may include 1 woman and 4 man such as ^{4}C_{1} x ^{6}C_{4} or committee may not include women and 5 men ^{6}C_{5}

(^{4}C_{2} x ^{6}C_{3})+(^{ 4}C_{1} x ^{6}C_{4}) + ^{6}C_{5} = 186

**A sports team of 11 students is to be constituted choosing at least 6 from class IX and at least 5 from class X. If there are 8 students in each of these classes, in how many ways can the team be constituted?**

Solution:

A sports team of 11 students is to be constituted by choosing in following way,

(i)6 students from class IX and 5 students from class X = ^{8}C_{6}*^{8}C_{5}

(ii)5 students from class IX and 6 students from class X = ^{8}C_{5}*^{8}C_{6}

= (^{8}C_{6}*^{8}C_{5}) + (^{8}C_{5}*^{8}C_{6})

= 1568 + 1568

= 3136

In 3136 ways a team can be constituted.

**From a group of 12 students, 8 are to be chosen for an excursion. There are 3 students who decide that either of them will join or none of them will join. In how many ways can the 8 be chosen?**

Solution:

From a group of 12 students, 8 are to be chosen for an excursion. There are 3 students who decide that either of them will join or none of them will join. If none of three students decide to join then ^{3}C_{0} x ^{9}C_{8} and if one of the student decide to join then ^{3}C_{1} X ^{9}C_{7}

Number ways 8 students can be chosen for excursion = ^{3}C_{0} x ^{9}C_{8} + ^{3}C_{1} x ^{9}C_{7}

= 1 x 9 + 3 x 36

= 9 + 108

= 117