10th mathematics exercise questions with answers, Mathematics

Permutation and Combination – Exercise 4.7 – Class X

Ashoora Arif

Previous  exercise –  Permutation and Combination – Exercise 4.6

Permutation and Combination – Exercise 4.7

  1. Out of 7 consonants and 4 towels, how many words of 3 consonants and 2 vowels can be formed?
  2. In how many ways can 5 sportsmen can be selected from a group of 10?
  3. 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?
  4. How many (i) lines (ii) triangles can be drawn through 8 points on a circle?
  5. How many diagonals can be drawn in a (i) decagon (ii) icosagon
  6. A polygon has 44 diagonals. Find the number of sides
  7. 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?

  1. A committee of 5 is to be formed out of 6 men and 4 ladies are included?
  2. 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?
  3. 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:

  1. 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 = (7C3 x 4C2) =  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

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

Solution:

Here n= 10 and r = 5

Then, nCr = 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.

  1. 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  = 12C9 x 5C2 =  2200

  1. 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 = 8C2 = 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 = 8C3 = 8!/(8-3)!3! = 8!/5!3! = 5!x6x7x8/5!x2x3 = 6x7x8/2×3 = 56

  1. 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 = nCr – n .

Here, n = 10 and r = 2

Thus, nCr – 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 = nCr – n .

Here, n = 20 and r = 2

Thus, nCr – n = n!/(n-r)!r! – n = 20!/(20-2)!2! – 10 = 18!x19x20/18!x1x2 – 10 = 190 – 20 = 170

  1. 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 = nCr – 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 = n2 – n – n

⇒88 = n2 – 2n

⇒n2 – 2n – 88 = 0

⇒n2 – 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.

  1. 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.

  1. 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 4C3 x 6C2 or committee may include 4 women and 1 man such as 4C4 x 6C1 or committee may include two women and 3 men 4C2 x 6C3

(4C3 x 6C2) + (4C4 x 6C1) + (4C2 x 6C3) = 186.

(ii) When at most 2 ladies are included i.e., committee may consist 2 women and 3 men such as 4C2 x 6C3 or committee may include 1 woman and 4 man such as 4C1 x 6C4 or committee may not include women and 5 men  6C5

(4C2 x 6C3)+( 4C1 x 6C4) + 6C5 = 186

  1. 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 =  8C6*8C5

(ii)5 students from class IX and 6 students from class X =  8C5*8C6

= (8C6*8C5) + (8C5*8C6)

= 1568 + 1568

= 3136

In 3136 ways a team can be constituted.

  1. 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 3C0 x 9C8 and if one of the student decide to join then 3C1 X 9C7

Number ways 8 students can be chosen for excursion  =  3C0 x 9C8 + 3C1 x 9C7

= 1 x 9  + 3 x 36

= 9 + 108

= 117

 

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Published by Ashoora Arif