CBSE Class 11 Mathematics Permutations and Combinations Competitive Exam MCQs Set 07

Answer: (b) 18

Explanation:
1. If there are n teams, the number of matches = \( {}^{n}C_{2} = 153 \).
2. \( n(n-1) = 306 = 18 \times 17 \Rightarrow n = 18 \).

Teacher's Note:
1. Matches correspond to pairs of teams.
2. Double the number of matches and look for consecutive factors.

Answer: (a) \( \frac{10!}{6} \)

Explanation:
1. Total orders = 10!.
2. A, B, C appear in 3! = 6 relative orders equally often; only A, B, C is wanted.
3. Required = \( \frac{10!}{6} \).

Teacher's Note:
1. Fixing the order of k persons divides the total by k!.
2. Equivalently choose 3 slots (¹⁰C₃) for A, B, C in that order and arrange the others in 7! ways.

Answer: (c) 300

Explanation:
1. Write numbers 000 to 999 as three digit strings: 3000 digits in all.
2. Each digit 0 to 9 appears equally often: 300 times.
3. 1000 has no 5, so the digit 5 is written 300 times.

Teacher's Note:
1. Padding with leading zeros makes every position equally likely for each digit.
2. 271 is the number of integers containing at least one 5, a different question.

Answer: (b) 7180

Explanation:
1. Total products (pairs of different numbers) = \( {}^{200}C_{2} = 19900 \).
2. Numbers not multiples of 5: 160; products with no multiple of 5: \( {}^{160}C_{2} = 12720 \).
3. Products that are multiples of 5 = 19900 − 12720 = 7180.

Teacher's Note:
1. A product is a multiple of 5 if at least one factor is a multiple of 5 (5 is prime).
2. Use complement: total pairs minus pairs from non-multiples.

Answer: (d) None of these

Explanation:
1. Driver's seat: 2 ways.
2. Other front seat and rear seat from the remaining 5 persons: \( {}^{5}P_{2} = 20 \) ways.
3. Total = 2 × 20 = 40, which is not in (a), (b), (c).

Teacher's Note:
1. The two remaining seats are different places, so the persons are arranged, not just selected.
2. Fill the driving seat first because only 2 persons qualify.

Answer: (b) 2⁸ − 2

Explanation:
1. Each bracket is \( {}^{8}C_{r+1} \) by Pascal's rule.
2. Sum = \( {}^{8}C_{1} + {}^{8}C_{2} + \ldots + {}^{8}C_{7} = 2^{8} - {}^{8}C_{0} - {}^{8}C_{8} = 2^{8} - 2 \).

Teacher's Note:
1. Combine each bracket first, then use the total 2ⁿ.
2. Remember to remove the missing end terms ⁸C₀ and ⁸C₈.

Answer: (a) \( {}^{n+4}C_{r} \)

Explanation:
1. Group as \( ({}^{n}C_{r} + {}^{n}C_{r-1}) + 3({}^{n}C_{r-1} + {}^{n}C_{r-2}) + 3({}^{n}C_{r-2} + {}^{n}C_{r-3}) + ({}^{n}C_{r-3} + {}^{n}C_{r-4}) \)
2. \( = {}^{n+1}C_{r} + 3\,{}^{n+1}C_{r-1} + 3\,{}^{n+1}C_{r-2} + {}^{n+1}C_{r-3} \)
3. Repeating the grouping: \( {}^{n+2}C_{r} + 2\,{}^{n+2}C_{r-1} + {}^{n+2}C_{r-2} = {}^{n+3}C_{r} + {}^{n+3}C_{r-1} = {}^{n+4}C_{r} \).

Teacher's Note:
1. The coefficients 1, 4, 6, 4, 1 come from Pascal's triangle, so Pascal's rule applied four times collapses the sum.
2. Each round reduces the coefficients to the previous row: 1, 3, 3, 1 → 1, 2, 1 → 1, 1.

Answer: (c) 63

Explanation:
1. Each friend is either invited or not: 2⁶ = 64 ways.
2. Excluding the case of inviting nobody: 64 − 1 = 63.

Teacher's Note:
1. Selections of one or more from n different things number 2ⁿ − 1.
2. This equals ⁶C₁ + ⁶C₂ + … + ⁶C₆.

Answer: (d) 63

Explanation:
1. Total ways of answering = 4³ = 64.
2. Only 1 way gets all answers correct.
3. Required = 64 − 1 = 63.

Teacher's Note:
1. 'Fail to get all correct' is the complement of 'all correct'.
2. Each question is answered independently in 4 ways.

Answer: (c) 218

Explanation:
1. Voter votes for 1 to 5 candidates: \( {}^{8}C_{1} + {}^{8}C_{2} + {}^{8}C_{3} + {}^{8}C_{4} + {}^{8}C_{5} \)
2. = 8 + 28 + 56 + 70 + 56 = 218.

Teacher's Note:
1. A vote is a selection of candidates, so combinations are used.
2. At least one vote is assumed, so ⁸C₀ is not counted.

Answer: (c) 8

Explanation:
1. Let n candidates, n − 1 to be elected. Ways = \( {}^{n}C_{1} + \ldots + {}^{n}C_{n-1} = 2^{n} - 2 \).
2. 2ⁿ − 2 = 254 ⇒ 2ⁿ = 256 ⇒ n = 8.

Teacher's Note:
1. The voter must vote for at least one and at most n − 1 candidates.
2. So remove ⁿC₀ and ⁿCₙ from 2ⁿ.

Answer: (c) 2³² − 1

Explanation:
1. Each of the 32 tooth positions is either filled or empty: 2³² patterns.
2. The pattern with no tooth is not allowed: 2³² − 1.
3. Since no two persons share a pattern, the maximum population is 2³² − 1.

Teacher's Note:
1. A set of teeth is a non-empty selection of the 32 positions.
2. Selections of one or more from n different things: 2ⁿ − 1.

Answer: (d) 2454

Explanation:
1. Letters: M, A, T (twice each) and H, E, I, C, S (once): 8 different letters.
2. All 4 different: \( {}^{8}C_{4} \times 4! = 1680 \). Two alike + two different: \( {}^{3}C_{1} \times {}^{7}C_{2} \times \frac{4!}{2!} = 756 \).
3. Two alike + two alike: \( {}^{3}C_{2} \times \frac{4!}{2!\,2!} = 18 \).
4. Total = 1680 + 756 + 18 = 2454.

Teacher's Note:
1. With repeated letters, split into cases by the pattern of repetition in the chosen 4 letters.
2. In each case, select the letters first and then arrange them, dividing for identical letters.

Answer: (a) 4

Explanation:
1. Ways = \( {}^{2n+1}C_{1} + \ldots + {}^{2n+1}C_{n} = S \).
2. By symmetry \( {}^{2n+1}C_{r} = {}^{2n+1}C_{2n+1-r} \), so \( 1 + 2S + 1 = 2^{2n+1} \Rightarrow S = 2^{2n} - 1 \).
3. 2²ⁿ − 1 = 255 ⇒ 2²ⁿ = 256 ⇒ n = 4.

Teacher's Note:
1. The coefficients of an odd row of Pascal's triangle split into two equal halves.
2. So each half (excluding the two end 1's) adds to 2²ⁿ − 1.

Answer: (a) 1540

Explanation:
1. Place the 21 English books; they create 22 gaps.
2. Choose 19 of the 22 gaps for the Hindi books: \( {}^{22}C_{19} = {}^{22}C_{3} = 1540 \).

Teacher's Note:
1. Since books of a language are alike, only the choice of gaps matters.
2. With different books, the answer would be multiplied by 21! × 19!.

Answer: (c) 9

Explanation:
1. Two black: \( {}^{4}C_{2} = 6 \); two white: \( {}^{3}C_{2} = 3 \).
2. Total = 6 + 3 = 9.

Teacher's Note:
1. The two cases are exclusive, so add them.
2. Balls are treated as distinguishable here.

Answer: (c) 35

Explanation:
1. The 6 persons not selected leave 7 gaps (including the ends).
2. Choose 4 of these 7 gaps for the selected persons: \( {}^{7}C_{4} = 35 \).

Teacher's Note:
1. Think of the unselected persons as separators.
2. Placing selected persons in different gaps ensures no two are neighbours.

Answer: (c) 96

Explanation:
1. Total words = \( {}^{6}P_{3} = 120 \).
2. Words with no vowel (from b, c, d, f) = \( {}^{4}P_{3} = 24 \).
3. Required = 120 − 24 = 96.

Teacher's Note:
1. 'At least one vowel' = total − none.
2. The vowels here are a and e.

Answer: (c) 100

Explanation:
1. 4 boys + 3 girls: \( {}^{6}C_{4} \times {}^{4}C_{3} = 60 \); 5 boys + 2 girls: \( {}^{6}C_{5} \times {}^{4}C_{2} = 36 \);
2. 6 boys + 1 girl: \( {}^{6}C_{6} \times {}^{4}C_{1} = 4 \).
3. Total = 60 + 36 + 4 = 100.

Teacher's Note:
1. Majority in a group of 7 means at least 4 boys.
2. List each case and add.

Answer: (b) 140

Explanation:
1. Total selections = \( {}^{10}C_{6} = 210 \).
2. Both those friends included: \( {}^{8}C_{4} = 70 \).
3. Required = 210 − 70 = 140.

Teacher's Note:
1. Remove the selections that contain both the friends.
2. Selections with only one of them or neither are allowed.

Answer: (b) 3⁷

Explanation:
1. Each of the 7 voters can vote for any of the 3 candidates.
2. Total = 3⁷.

Teacher's Note:
1. The voters make the choices, so the number of candidates is the base.
2. 7³ would wrongly let candidates choose voters.

Answer: (c) 90

Explanation:
1. Choose which boy gets 2 balls: 2 ways.
2. Choose those 2 balls: \( {}^{10}C_{2} = 45 \); the other boy gets the rest.
3. Total = 2 × 45 = 90.

Teacher's Note:
1. The boys are different, so who gets the smaller share matters.
2. 45 would only count the ways of splitting the balls into a 2-group and an 8-group.

Answer: (a) \( {}^{52}C_{26} \cdot 2^{26} \)

Explanation:
1. Choose 26 different cards (by suit and denomination) out of 52 kinds: \( {}^{52}C_{26} \).
2. Each chosen kind can come from either pack: 2 ways each, i.e., 2²⁶.
3. Total = \( {}^{52}C_{26} \cdot 2^{26} \).

Teacher's Note:
1. Each card type appears twice in the combined pack.
2. Select the types first, then the copy of each.

Answer: (c) 13

Explanation:
1. Let there be m men. Games among men = \( 2 \times {}^{m}C_{2} = m(m-1) \); games of men with women = 2 × 2m = 4m.
2. m(m − 1) − 4m = 66 ⇒ m² − 5m − 66 = 0 ⇒ (m − 11)(m + 6) = 0 ⇒ m = 11.
3. Participants = 11 + 2 = 13.

Teacher's Note:
1. Every pair plays two games, so multiply the number of pairs by 2.
2. Do not forget to add the two women at the end.

Answer: (a) 64

Explanation:
1. Total ways = \( {}^{9}C_{3} = 84 \).
2. No black ball: \( {}^{6}C_{3} = 20 \).
3. Required = 84 − 20 = 64.

Teacher's Note:
1. 'At least one' problems are quickest by complement.
2. Balls are treated as distinct objects.

Answer: (b) 246

Explanation:
1. Total = \( {}^{10}C_{5} = 252 \).
2. All men = \( {}^{6}C_{5} = 6 \).
3. Required = 252 − 6 = 246.

Teacher's Note:
1. Subtract the committees without any woman.
2. Direct case-wise addition gives the same result.

Answer: (c) 35

Explanation:
1. The six ‘+’ signs create 7 gaps.
2. Choose 4 gaps for the ‘−’ signs: \( {}^{7}C_{4} = 35 \).

Teacher's Note:
1. Identical signs mean only the choice of gaps matters.
2. Gap method: arrange the unrestricted objects first.

Answer: (b) 3720

Explanation:
1. Green: 2⁵ − 1 = 31 ways; blue: 2⁴ − 1 = 15 ways; red (any number): 2³ = 8 ways.
2. Total = 31 × 15 × 8 = 3720.

Teacher's Note:
1. For each colour, count the subsets and remove the empty one when at least one is required.
2. Red balls have no restriction, so the empty choice is allowed.

Answer: (d) 25

Explanation:
1. The voter may vote for 1, 2 or 3 of the 5 candidates.
2. Ways = \( {}^{5}C_{1} + {}^{5}C_{2} + {}^{5}C_{3} = 5 + 10 + 10 = 25 \).

Teacher's Note:
1. The voter must vote for at least one and at most three candidates.
2. Each vote is a selection, so use combinations.

Answer: (d) 2702, 1008

Explanation:
1. Women majority (7, 8 or 9 women): \( {}^{9}C_{7}{}^{8}C_{5} + {}^{9}C_{8}{}^{8}C_{4} + {}^{9}C_{9}{}^{8}C_{3} = 2016 + 630 + 56 = 2702 \).
2. Men majority with at least 5 women means exactly 5 women, 7 men: \( {}^{9}C_{5}{}^{8}C_{7} = 126 \times 8 = 1008 \).

Teacher's Note:
1. The condition 'at least 5 women' still applies when men are in majority.
2. With 6 women and 6 men there is no majority.

Answer: (b) \( 2({}^{8}C_{4}) \)

Explanation:
1. Put the two particular persons in different boats: 2 ways.
2. Choose 4 of the remaining 8 to join the first of them: \( {}^{8}C_{4} \); the rest go in the other boat.
3. Total = \( 2({}^{8}C_{4}) = 140 \).

Teacher's Note:
1. The boats are different, so swapping the two particular persons gives a new way.
2. Once one boat is filled, the other is determined.

Answer: (d) \( {}^{7}C_{3}(5!) \)

Explanation:
1. A is included, G and H excluded: choose 4 more from the remaining 7: \( {}^{7}C_{4} = {}^{7}C_{3} \).
2. Arrange the 5 in a line: 5!. Total = \( {}^{7}C_{3}(5!) \).

Teacher's Note:
1. Handle inclusion and exclusion first, then arrange.
2. ⁷C₄ = ⁷C₃ = 35.

Answer: (b) 3605

Explanation:
1. All selections: \( {}^{30}C_{3} = 4060 \); all-even selections: \( {}^{15}C_{3} = 455 \).
2. Required = 4060 − 455 = 3605.

Teacher's Note:
1. There are 15 even numbers from 1 to 30.
2. Subtract the unwanted selections from the total.

Answer: (b) 3¹²

Explanation:
1. Each of the 12 stalls can be filled with a horse, a cow or a calf: 3 ways.
2. Total = 3¹².

Teacher's Note:
1. Animals of the same kind are treated as alike, so only the kind in each stall matters.
2. Having at least 12 of each kind means every pattern is possible.

Answer: (d) 2{(n + 1)!}²

Explanation:
1. Colours must alternate: WBWB… or BWBW…: 2 patterns.
2. In each pattern, white balls are arranged in (n + 1)! ways and black balls in (n + 1)! ways.
3. Total = 2{(n + 1)!}².

Teacher's Note:
1. Equal numbers of two colours alternating can start with either colour.
2. The balls are numbered, so they are distinct within a colour.

Answer: (b) 16³ × 15² × 14

Explanation:
1. Running: 16 ways.
2. Swimming (2 different prizes): 16 × 15.
3. Riding (3 different prizes): 16 × 15 × 14.
4. Total = 16³ × 15² × 14.

Teacher's Note:
1. Within one event a man cannot win two prizes, but he may win in different events.
2. Multiply the independent counts for the three events.

Answer: (a) 252

Explanation:
1. 3 ladies: \( {}^{4}C_{3} \times {}^{8}C_{3} = 224 \); 4 ladies: \( {}^{4}C_{4} \times {}^{8}C_{2} = 28 \).
2. Total = 252.

Teacher's Note:
1. Only 4 ladies are available, so the cases are 3 or 4 ladies.
2. Add the cases.

Answer: (b) 196

Explanation:
1. 4 from first five and 6 from the other 8: \( {}^{5}C_{4} \times {}^{8}C_{6} = 140 \).
2. 5 from first five and 5 from the other 8: \( {}^{5}C_{5} \times {}^{8}C_{5} = 56 \).
3. Total = 196.

Teacher's Note:
1. Split by the number chosen from the first five.
2. Only 8 questions remain in the second group.

Answer: (b) 53

Explanation:
1. Group stage: \( 2 \times {}^{6}C_{2} = 30 \).
2. Next round (6 teams): \( {}^{6}C_{2} = 15 \).
3. Semifinal round (4 teams): \( {}^{4}C_{2} = 6 \).
4. Final (best of three): at least 2. Total = 30 + 15 + 6 + 2 = 53.

Teacher's Note:
1. Each round-robin stage of k teams has ᵏC₂ matches.
2. The minimum for a best-of-three is 2 matches.

Answer: (a) 485

Explanation:
1. If the man invites k women and 3 − k men, the wife invites 3 − k women and k men.
2. k = 0: 1 × 1 = 1; k = 1: (4 × 3)(3 × 4) = 144; k = 2: (6 × 3)(3 × 6) = 324; k = 3: 4 × 4 = 16.
3. Total = 1 + 144 + 324 + 16 = 485.

Teacher's Note:
1. Fix the man's choice and the wife's choice is forced to make 3 women and 3 men overall.
2. Write each case as a product of four combination counts.

Answer: (b) 10

Explanation:
1. With r friends at a time, the number of different parties is \( {}^{20}C_{r} \).
2. For even n, \( {}^{n}C_{r} \) is greatest at r = n/2, so r = 10.

Teacher's Note:
1. For even n, ⁿCᵣ is greatest at r = n/2.
2. So invite 10 friends at a time.

Answer: (c) \( {}^{8}C_{3} - {}^{5}C_{3} - 1 \)

Explanation:
1. All selections of 3 points: \( {}^{8}C_{3} \).
2. Collinear triples: \( {}^{5}C_{3} \) on one line and \( {}^{3}C_{3} = 1 \) on the other.
3. Triangles = \( {}^{8}C_{3} - {}^{5}C_{3} - 1 = 45 \).

Teacher's Note:
1. Remove collinear triples from both lines.
2. Three points on the second line also fail to form a triangle.