RD Chapter 17 Combinations Ex 17.2 Solutions
Question - 11 : - A student has to answer 10 questions, choosing at least 4 from each of part A and part B. If there are 6 questions in part A and 7 in part B, in how many ways can the student choose 10 questions?
Answer - 11 : -
Given:
Total number ofquestions = 10
Questions in part A =6
Questions in part B =7
Number of ways = (No.of ways of answering 4 questions from part A and 6 from part B) + (No. of waysof answering 5 questions from part A and 5 questions from part B) + (No. ofways of answering 6 questions from part A and 4 from part B)
= (6C4 × 7C6)+ (6C5 × 7C5) + (6C6 × 7C4)
By using the formula,
nCr = n!/r!(n – r)!
= (15×7) + (6×21) +(1×35)
= 105 + 126 + 35
= 266
∴ The total no. ofways of answering 10 questions is 266 ways.
Question - 12 : - In an examination, a student to answer 4 questions out of 5 questions; questions 1 and 2 are however compulsory. Determine the number of ways in which the student can make a choice.
Answer - 12 : -
Given:
Total number ofquestions = 5
Total number ofquestions to be answered = 4
Number of ways = weneed to answer 2 questions out of the remaining 3 questions as 1 and 2 arecompulsory.
= 3C2
By using the formula,
nCr = n!/r!(n – r)!
3C2 = 3!/2!(3 -2)!
= 3! / (2! 1!)
= [3×2×1] / (2×1)
= 3
∴ The no. of waysanswering the questions is 3.
Question - 13 : - A candidate is required to answer 7 questions out of 12 questions which are divided into two groups, each containing 6 questions. He is not permitted to attempt more than 5 questions from either group. In how many ways can he choose the 7 questions?
Answer - 13 : -
Given:
Total number ofquestions = 12
Total number ofquestions to be answered = 7
Number of ways = (No.of ways of answering 5 questions from group 1 and 2 from group 2) + (No. ofways of answering 4 questions from group 1 and 3 from group 2) + (No. of waysof answering 3 questions from group 1 and 4 from group 2) + (No. of ways ofanswering 2 questions from group 1 and 5 from group 2)
= (6C5 × 6C2)+ (6C4 × 6C3) + (6C3 × 6C4)+ (6C2 × 6C5)
By using the formula,
nCr = n!/r!(n – r)!
= (6×15) + (15×20) +(20×15) + (15×6)
= 90 + 300 + 300 + 90
= 780
∴ The total no. ofways of answering 7 questions is 780 ways.
Question - 14 : - There are 10 points in a plane of which 4 are collinear. How many different straight lines can be drawn by joining these points.
Answer - 14 : -
Given:
Total number of points= 10
Number of collinearpoints = 4
Number of lines formed= (total no. of lines formed by all 10 points) – (no. of lines formed bycollinear points) + 1
Here, 1 is addedbecause only 1 line can be formed by the four collinear points.
= 10C2 – 4C2 +1
By using the formula,
nCr = n!/r!(n – r)!
= 90/2 – 12/2 + 1
= 45 – 6 + 1
= 40
∴ The total no. ofways of different lines formed are 40.
Question - 15 : - Find the number of diagonals of
(i) a hexagon
(ii) a polygon of 16 sides
Answer - 15 : -
(i) a hexagon
We know that a hexagonhas 6 angular points. By joining those any two angular points we get a linewhich is either a side or a diagonal.
So number of linesformed = 6C2
By using the formula,
nCr = n!/r!(n – r)!
6C2 = 6!/2!(6-2)!
= 6! / (2! 4!)
= [6×5×4!] / (2! 4!)
= [6×5] / (2×1)
= 3×5
= 15
We know number ofsides of hexagon is 6
So, number of diagonals= 15 – 6 = 9
The total no. ofdiagonals formed is 9.
(ii) a polygon of 16 sides
We know that a polygonof 16 sides has 16 angular points. By joining those any two angular points weget a line which is either a side or a diagonal.
So number of linesformed = 16C2
By using the formula,
nCr = n!/r!(n – r)!
16C2 = 16!/2!(16-2)!
= 16! / (2! 14!)
= [16×15×14!] / (2!14!)
= [16×15] / (2×1)
= 8×15
= 120
We know number ofsides of a polygon is 16
So, number ofdiagonals = 120 – 16 = 104
The total no. ofdiagonals formed is 104.
Question - 16 : - How many triangles can be obtained by joining 12 points, five of which are collinear?
Answer - 16 : -
We know that 3 points are required to draw a triangle and the collinear points will lie on the same line.
Number of triangles formed = (total no. of triangles formed by all 12 points) – (no. of triangles formed by collinear points)
= 12C3 – 5C3
By using the formula,
nCr = n!/r!(n – r)!
= (2×11×10) – (5×2)
= 220 – 10
= 210
∴ The total no. oftriangles formed are 210.