Balbharati Maharashtra State Board11th Commerce Maths Solution Book PdfChapter 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
Maharashtra Board Solution
Find the value of (i) 15C4
Solution & Step-by-Step Answer:
(ii)80C2
Solution:
(iii)15C4+15C5
Solution:
(iv)20C16–19C16
Solution:
Question 2
Maharashtra Board Solution
Find n if (i) 6P2 = n 6C2
Solution & Step-by-Step Answer:
(ii)2nC3:nC2= 52 : 3
Solution:
(iii)nCn-3= 84
Solution:
nCn-3= 84
∴ = 84
∴ = 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
Maharashtra Board Solution
Find r if 14C2r : 10C2r-4 = 143 : 10
Solution & Step-by-Step Answer:
∴ ∴ 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
Maharashtra Board Solution
Find n and r if. (i) nPr = 720 and nCn-r = 120
Solution & Step-by-Step Answer:
(ii)nCr-1:nCr:nCr+1= 20 : 35 : 42
Solution:
Question 5
Maharashtra Board Solution
If nPr = 1814400 and nCr = 45, find r.
Solution & Step-by-Step Answer:
∴ r! = 40320 ∴ r! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 ∴ r! = 8! ∴ r = 8
Question 6
Maharashtra Board Solution
If nCr-1 = 6435, nCr = 5005, nCr+1 = 3003, find rC5.
Solution & Step-by-Step Answer:
Question 7
Maharashtra Board Solution
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 & Step-by-Step Answer:
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 6C3 ways. 3 green balls can be selected from 5 green balls in 5C3 ways. 3 blue balls can be selected from 7 blue balls in 7C3 ways. ∴ Number of ways selection can be done if the selection consists of 3 balls of each colour
Question 8
Maharashtra Board Solution
Find the number of ways of selecting a team of 3 boys and 2 girls from 6 boys and 4 girls.
Solution & Step-by-Step Answer:
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 6C3 ways. 2 girls can be selected from 4 girls in 4C2 ways. ∴ Number of ways the team can be selected = 6C3 × 4C2 = = = 20 × 6 = 120 ∴ The team of 3 boys and 2 girls can be selected in 120 ways.
Question 9
Maharashtra Board Solution
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 & Step-by-Step Answer:
Let there be n participants present in the meeting. A handshake occurs between 2 persons. ∴ Number of handshakes = nC2 Given 66 handshakes were exchanged. ∴ 66 = nC2 ∴ 66 = ∴ 66 × 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
Maharashtra Board Solution
If 20 points are marked on a circle, how many chords can be drawn?
Solution & Step-by-Step Answer:
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 = 20C2 = = = 190
Question 11
Maharashtra Board Solution
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 & Step-by-Step Answer:
In n-sided polygon, there are ‘n’ points and ‘n’ sides.. ∴ Through ‘n’ points we can draw nC2 lines including sides. ∴ Number of diagonals in n sided polygon = nC2 – n (∴ n = number of sides)
Question 12
Maharashtra Board Solution
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 & Step-by-Step Answer:
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 = 20C2 = = = 190
Question 13
Maharashtra Board Solution
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 & Step-by-Step Answer:
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 = 10C2 = = = 5 × 9 = 45
(ii) When 4 of them arc collinear:
∴ Number of lines passing through these points if 4 points are collinear
=10C2–4C2+ 1
= 45 – + 1
= 45 – + 1
= 45 – 6 + 1
= 40
Question 14
Maharashtra Board Solution
Find the number of triangles formed by joining 12 points if (i) no three points are collinear (ii) four points are collinear
Solution & Step-by-Step Answer:
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 = 12C3 = = = 220
(ii) When 4 of these points are collinear:
∴ Number of triangles that can be obtained from these points =12C3–4C3
= 220 –
= 220 –
= 220 – 4
= 216
Question 15
Maharashtra Board Solution
A word has 8 consonants and 3 vowels. How many distinct words can be formed if 4 consonants and 2 vowels are chosen?
Solution & Step-by-Step Answer:
Out of 8 consonants, 4 can be selected in 8C4 = = = 70 ways From 3 vowels, 2 can be selected in 3C2 = = = 3 ways Now, to form a word, these 6 letters (i.e., 4 consonants and 2 vowels) can be arranged in 6P6 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.