CBSE Class 11 Mathematics Permutations and Combinations Competitive Exam MCQs Set 08

Answer: (a) 190

Explanation:
1. Two straight lines meet in at most one point; the maximum occurs when no two are parallel and no three are concurrent.
2. Points = number of pairs of lines = \( {}^{20}C_{2} = 190 \).

Teacher's Note:
1. Maximum means no two lines parallel and no three concurrent.
2. Each point then corresponds to exactly one pair of lines.

Answer: (b) 11

Explanation:
1. \( {}^{n}C_{2} - n = \frac{n(n-3)}{2} = 44 \Rightarrow n(n-3) = 88 = 11 \times 8 \)
2. n = 11.

Teacher's Note:
1. Diagonals = lines joining vertices minus the sides.
2. Solve n(n − 3) = 2 × diagonals.

Answer: (a) 84

Explanation:
1. No three of the points are collinear, so any 3 points form a triangle.
2. Triangles = \( {}^{9}C_{3} = 84 \).

Teacher's Note:
1. 'Non-collinear' here means no three points are on a line.
2. A triangle is a selection of 3 vertices.

Answer: (c) \( \frac{1}{2!}m(m-3) \)

Explanation:
1. Lines joining pairs of vertices = \( {}^{m}C_{2} \); m of these are sides.
2. Diagonals = \( {}^{m}C_{2} - m = \frac{m(m-1)}{2} - m = \frac{m(m-3)}{2} \).

Teacher's Note:
1. Every pair of vertices gives a side or a diagonal.
2. Check: a hexagon has 6 × 3/2 = 9 diagonals.

Answer: (b) 116

Explanation:
1. All triples of points: \( {}^{10}C_{3} = 120 \); triples from the 4 collinear points: \( {}^{4}C_{3} = 4 \) do not form triangles.
2. Triangles = 120 − 4 = 116.

Teacher's Note:
1. Three collinear points do not form a triangle.
2. Subtract all triples taken from the collinear points.

Answer: (a) 106

Explanation:
1. Pairs of points: \( {}^{16}C_{2} = 120 \); the 6 collinear points give \( {}^{6}C_{2} = 15 \) pairs but only one line.
2. Lines = 120 − 15 + 1 = 106.

Teacher's Note:
1. The 6 collinear points give only one line, not ⁶C₂ lines.
2. So subtract ⁶C₂ and add 1.

Answer: (b) 18

Explanation:
1. A parallelogram is formed by choosing 2 lines from each set.
2. Number = \( {}^{4}C_{2} \times {}^{3}C_{2} = 6 \times 3 = 18 \).

Teacher's Note:
1. A parallelogram needs two lines from each family.
2. Multiply the selections from the two families.

Answer: (d) 104

Explanation:
1. Line–line: \( {}^{8}C_{2} = 28 \); circle–circle: \( 2 \times {}^{4}C_{2} = 12 \); line–circle: 2 × 8 × 4 = 64.
2. Total = 28 + 12 + 64 = 104.

Teacher's Note:
1. Two lines meet in at most 1 point, two circles or a line and a circle in at most 2.
2. Add the three types of intersections.

Answer: (a) 504

Explanation:
1. All triples: \( {}^{16}C_{3} = 560 \); triples from the 8 collinear points: \( {}^{8}C_{3} = 56 \).
2. Triangles = 560 − 56 = 504.

Teacher's Note:
1. Subtract the triples taken from the 8 collinear points.
2. All other triples form triangles.

Answer: (b) 7

Explanation:
1. \( T_{n} = {}^{n}C_{3} \). \( T_{n+1} - T_{n} = {}^{n+1}C_{3} - {}^{n}C_{3} = {}^{n}C_{2} \).
2. \( {}^{n}C_{2} = 21 \Rightarrow n(n-1) = 42 \Rightarrow n = 7 \).

Teacher's Note:
1. Pascal's rule gives the difference directly.
2. No three vertices of a polygon are collinear.

Answer: (a) 100

Explanation:
1. All triples: \( {}^{10}C_{3} = 120 \); triples from the 6 collinear points: \( {}^{6}C_{3} = 20 \).
2. Triangles = 120 − 20 = 100.

Teacher's Note:
1. Remove all triples from the collinear points.
2. Only those triples fail to form triangles.

Answer: (d) 185

Explanation:
1. Pairs of points: \( {}^{20}C_{2} = 190 \); the 4 collinear points give \( {}^{4}C_{2} = 6 \) pairs but only one line.
2. Lines = 190 − 6 + 1 = 185.

Teacher's Note:
1. The 4 collinear points give 1 line instead of 6.
2. Use ⁿC₂ − ᵐC₂ + 1.

Answer: (a) \( {}^{6}C_{3} - 7 \)

Explanation:
1. A triangle needs the sum of the two smaller sides to exceed the largest side.
2. Triples that fail: (2, 3, 5), (2, 3, 6), (2, 3, 7), (2, 4, 6), (2, 4, 7), (2, 5, 7), (3, 4, 7): 7 triples.
3. Triangles = \( {}^{6}C_{3} - 7 = 13 \).

Teacher's Note:
1. Check each triple against the triangle inequality, especially those containing 2 or 3.
2. A triple like (2, 3, 5) with equality also fails.

Answer: (c) 10

Explanation:
1. Diagonals of an n-sided polygon = \( \frac{n(n-3)}{2} = 35 \).
2. n(n − 3) = 70 = 10 × 7 ⇒ n = 10.

Teacher's Note:
1. Use diagonals = n(n − 3)/2.
2. Look for factors of 70 differing by 3.

Answer: (b) 60

Explanation:
1. A parallelogram is formed by 2 lines from each family.
2. Number = \( {}^{5}C_{2} \times {}^{4}C_{2} = 10 \times 6 = 60 \).

Teacher's Note:
1. Each parallelogram uses two lines of each family.
2. Multiply the two selections.

Answer: (d) 56

Explanation:
1. Number of pairs of circles = \( {}^{8}C_{2} = 28 \).
2. Two circles meet in at most 2 points, so the maximum = 2 × 28 = 56.

Teacher's Note:
1. Count pairs of circles and multiply by the maximum intersections per pair.
2. For lines the multiplier would be 1.

Answer: (c) 117

Explanation:
1. Every 3 points determine a circle: \( {}^{10}C_{3} = 120 \).
2. The 4 concyclic points give \( {}^{4}C_{3} = 4 \) triples but only 1 circle.
3. Circles = 120 − 4 + 1 = 117.

Teacher's Note:
1. This is like the collinear-points formula for lines.
2. Three non-collinear points determine exactly one circle.

Answer: (a) 205

Explanation:
1. Total triples from 12 points: \( {}^{12}C_{3} = 220 \).
2. Collinear triples: \( {}^{3}C_{3} + {}^{4}C_{3} + {}^{5}C_{3} = 1 + 4 + 10 = 15 \).
3. Triangles = 220 − 15 = 205.

Teacher's Note:
1. Points on the same side are collinear.
2. The vertices A, B, C are not among the chosen points.

Answer: (c) 26

Explanation:
1. The figure has 8 squares (rows of 2, 4 and 2). Ways to choose 6 squares: \( {}^{8}C_{6} = 28 \).
2. A row is empty only if the 6 crosses fill all the other 6 squares; this can happen for the top row (1 way) or the bottom row (1 way). The middle row cannot be empty.
3. Required = 28 − 2 = 26.

Teacher's Note:
1. Count all placements and subtract those leaving a row empty.
2. Only a 2-square row can be left empty, since 6 crosses need 6 other squares.

Answer: (b) \( {}^{m+n+k}C_{3} - {}^{m}C_{3} - {}^{n}C_{3} - {}^{k}C_{3} \)

Explanation:
1. All triples: \( {}^{m+n+k}C_{3} \).
2. Collinear triples lie entirely on one line: ᵐC₃ + ⁿC₃ + ᵏC₃.
3. Triangles = \( {}^{m+n+k}C_{3} - {}^{m}C_{3} - {}^{n}C_{3} - {}^{k}C_{3} \).

Teacher's Note:
1. Points on different parallel lines are never collinear with points of a single line.
2. Subtract the collinear triples of each line.

Answer: (c) 51

Explanation:
1. Lines = \( {}^{6}C_{2} = 15 \); pairs of lines = \( {}^{15}C_{2} = 105 \).
2. At each original point 5 lines meet; their \( {}^{5}C_{2} = 10 \) pairs give the same point. So the 6 points account for 6 × 10 = 60 pairs.
3. New points = 105 − 60 = 45; together with the 6 original points, distinct points = 51.

Teacher's Note:
1. Pairs of lines through a common original point all meet at that point.
2. Count the original 6 points once each.

Answer: (a) \( \frac{m+n-2}{m+n} \)

Explanation:
1. (i) Triangles use 2 points on one line and 1 on the other: \( {}^{m}C_{2}\,n + {}^{n}C_{2}\,m = \frac{mn(m+n-2)}{2} \).
2. (ii) Including A adds mn more triangles (A with one point on each line): total = mn(m + n)/2.
3. Ratio = \( \frac{m+n-2}{m+n} \).

Teacher's Note:
1. A triangle needs points on both lines.
2. A lies on both lines, so it pairs with one point from each.

Answer: (c) \( \frac{n(n-1)(n-2)(n-3)}{8} \)

Explanation:
1. Each point of intersection lies on exactly two of the original lines. A fresh line joins two points that do not lie on a common original line.
2. Such a pair of points uses 4 different original lines, split into two pairs. Choose the 4 lines in \( {}^{n}C_{4} \) ways and split them into two pairs in 3 ways.
3. Fresh lines = \( 3\,{}^{n}C_{4} = \frac{n(n-1)(n-2)(n-3)}{8} \).

Teacher's Note:
1. A fresh line joins two points that are not on the same original line.
2. Equivalently, choose 4 original lines (ⁿC₄) and pair them into two pairs (3 ways): 3 × ⁿC₄.

Answer: (c) \( ({}^{m+2}C_{2})^{2} \)

Explanation:
1. Each direction now has m + 2 lines (the m lines and two sides).
2. Parallelograms = \( {}^{m+2}C_{2} \times {}^{m+2}C_{2} = ({}^{m+2}C_{2})^{2} \).

Teacher's Note:
1. Do not forget the sides of the original parallelogram.
2. Choose two lines in each direction.

Answer: (a) 535

Explanation:
1. If all were in general position: \( {}^{37}C_{2} = 666 \) points.
2. The 13 lines through A give \( {}^{13}C_{2} = 78 \) pairs but only 1 point; the 11 lines through B give 55 pairs but 1 point.
3. Points = 666 − 78 − 55 + 2 = 535.

Teacher's Note:
1. Concurrent lines meet in a single point instead of many.
2. Subtract the pairs and add back one point for each concurrency.

Answer: (c) \( {}^{n}C_{2} - {}^{p}C_{2} + 1 \)

Explanation:
1. Joining all pairs gives \( {}^{n}C_{2} \) lines if no three points were collinear.
2. The p collinear points give \( {}^{p}C_{2} \) pairs but only one line, so lines = \( {}^{n}C_{2} - {}^{p}C_{2} + 1 \).

Teacher's Note:
1. Replace the ᵖC₂ lines among collinear points by the single line through them.
2. This is a standard result for geometric counting.

Answer: (d) 120

Explanation:
1. Angular points of a decagon are its 10 vertices, no three of which are collinear.
2. Triangles = \( {}^{10}C_{3} = 120 \).

Teacher's Note:
1. No three vertices of a convex polygon are collinear.
2. Angular points means vertices.

Answer: (d) 16

Explanation:
1. Total triangles = \( {}^{8}C_{3} = 56 \).
2. Triangles with two sides of the octagon (three consecutive vertices): 8. Triangles with exactly one side: 8 × 4 = 32.
3. Required = 56 − 8 − 32 = 16.

Teacher's Note:
1. For exactly one side, choose the side (8 ways) and a third vertex not adjacent to it (8 − 4 = 4 ways).
2. Alternatively the formula n(n − 4)(n − 5)/6 gives 8 × 4 × 3/6 = 16.

Answer: (a) 20

Explanation:
1. Each interior intersection point comes from 4 vertices: \( {}^{n}C_{4} = 70 \Rightarrow n = 8 \).
2. Diagonals = \( {}^{8}C_{2} - 8 = 20 \).

Teacher's Note:
1. Any 4 vertices form a convex quadrilateral whose diagonals meet once inside.
2. Then use ⁿC₂ − n for the diagonals.

Answer: (c) \( {}^{n}C_{2} \times {}^{n}C_{2} \)

Explanation:
1. Two points on each line form a quadrilateral whose two crossing segments meet once between the lines.
2. Number of such points = \( {}^{n}C_{2} \times {}^{n}C_{2} \) (assuming no three segments are concurrent).

Teacher's Note:
1. Every intersection point is determined by 2 points on each line.
2. The maximum count assumes no three segments pass through one point.

Answer: (d) \( \frac{mn(m-1)(n-1)}{4} \)

Explanation:
1. A parallelogram needs 2 lines from each family.
2. \( {}^{m}C_{2} \times {}^{n}C_{2} = \frac{m(m-1)}{2} \cdot \frac{n(n-1)}{2} = \frac{mn(m-1)(n-1)}{4} \).

Teacher's Note:
1. Choose two lines from each family.
2. Then simplify the product.

Answer: (c) p²(4p − 3)

Explanation:
1. Total triples: \( {}^{3p}C_{3} \); collinear triples: \( 3\,{}^{p}C_{3} \).
2. \( {}^{3p}C_{3} - 3\,{}^{p}C_{3} = \frac{3p(3p-1)(3p-2) - 3p(p-1)(p-2)}{6} = p^{2}(4p-3) \).

Teacher's Note:
1. Subtract triples from the same line.
2. Check with p = 2: ⁶C₃ − 0 = 20 = 4 × 5.

Answer: (a) 6

Explanation:
1. \( {}^{n}P_{r} = r! \times {}^{n}C_{r} \), so \( r! = \frac{{}^{n}P_{r}}{{}^{n}C_{r}} = 720 \).
2. 720 = 6!, so r = 6.

Teacher's Note:
1. The ratio ⁿPᵣ/ⁿCᵣ is always r!.
2. Recognise 720 = 6!.

Answer: (c) 9

Explanation:
1. \( \frac{n!}{(n-4)!} = 24 \cdot \frac{n!}{5!\,(n-5)!} \Rightarrow \frac{1}{n-4} = \frac{24}{120} \)
2. n − 4 = 5 ⇒ n = 9.

Teacher's Note:
1. Write (n − 4)! = (n − 4)(n − 5)! to cancel.
2. Check: ⁹P₄ = 3024 and 24 × ⁹C₅ = 24 × 126 = 3024.

Answer: (a) 5

Explanation:
1. \( n(n-1)(n-2) + \frac{n(n-1)}{2} = 14n \Rightarrow (n-1)(n-2) + \frac{n-1}{2} = 14 \)
2. 2n² − 5n − 25 = 0 ⇒ (n − 5)(2n + 5) = 0 ⇒ n = 5.

Teacher's Note:
1. ⁿCₙ₋₂ = ⁿC₂ = n(n − 1)/2.
2. Divide by n first to reduce the degree.

Answer: (c) 8

Explanation:
1. \( \frac{n!}{(n-4)!} = 30 \cdot \frac{n!}{5!\,(n-5)!} \Rightarrow \frac{1}{n-4} = \frac{30}{120} \)
2. n − 4 = 4 ⇒ n = 8.

Teacher's Note:
1. Cancel n! and (n − 5)! from both sides.
2. Check: ⁸P₄ = 1680 = 30 × 56.

Answer: (d) 7

Explanation:
1. r! = 840/35 = 24 ⇒ r = 4.
2. ⁿP₄ = n(n − 1)(n − 2)(n − 3) = 840 = 7 × 6 × 5 × 4 ⇒ n = 7.

Teacher's Note:
1. Divide to find r! first.
2. Then match ⁿPᵣ with a product of consecutive integers.

Answer: (a) 3

Explanation:
1. ⁿCᵣ = ⁿCᵣ₋₁ ⇒ r + (r − 1) = n ⇒ n = 2r − 1.
2. ⁿPᵣ₊₁ = ⁿPᵣ (n − r) ⇒ n − r = 1.
3. So 2r − 1 = r + 1 ⇒ r = 2, n = 3.

Teacher's Note:
1. ⁿPᵣ₊₁ = ⁿPᵣ × (n − r).
2. Solve the two linear equations together.

Answer: (d) r!

Explanation:
1. \( {}^{n}P_{r} = \frac{n!}{(n-r)!} \) and \( {}^{n}C_{r} = \frac{n!}{r!\,(n-r)!} \).
2. \( {}^{n}P_{r} \div {}^{n}C_{r} = r! \).

Teacher's Note:
1. Each selection of r objects gives r! arrangements.
2. This is the basic relation between permutations and combinations.

Answer: (a) 7200

Explanation:
1. First digit: 9 ways; second and third: 10 ways each.
2. Last digit not 0 or 5: 8 ways.
3. Total = 9 × 10 × 10 × 8 = 7200.

Teacher's Note:
1. A number is divisible by 5 only when its last digit is 0 or 5.
2. Repetition is allowed since nothing forbids it.

Answer: (d) 2²ⁿ

Explanation:
1. Required = \( {}^{2n+1}C_{0} + {}^{2n+1}C_{1} + \ldots + {}^{2n+1}C_{n} \).
2. By \( {}^{2n+1}C_{r} = {}^{2n+1}C_{2n+1-r} \), this is exactly half of \( 2^{2n+1} \).
3. So the number of subsets = 2²ⁿ.

Teacher's Note:
1. Subsets with at most n elements pair off with their complements, which have at least n + 1 elements.
2. So exactly half of all 2²ⁿ⁺¹ subsets qualify.

Answer: (b) a₁ + a₂ + a₃ + … + a_k

Explanation:
1. Students with exactly i wrong answers = aᵢ − aᵢ₊₁ (with aₖ₊₁ = 0).
2. Total wrong = 1(a₁ − a₂) + 2(a₂ − a₃) + … + k aₖ.
3. = a₁ + a₂ + … + aₖ.

Teacher's Note:
1. Convert 'at least i' counts to 'exactly i' counts by subtraction.
2. The sum telescopes neatly.