CBSE Class 11 Mathematics Permutations and Combinations Competitive Exam MCQs Set 02

Answer: (a) 672

Explanation:
1. Choose the 2 positions for the digit 2: \( {}^{7}C_{2} \) = 21 ways.
2. Each of the other 5 positions is 1 or 3: 2⁵ = 32 ways.
3. Total = 21 × 32 = 672

Teacher's Note:
1. Place the restricted digit first, then fill the rest freely.
2. Since the digits are 1, 2, 3 only, no number can begin with 0.

Answer: (c) 15

Explanation:
1. Only the choice of positions matters: choose 2 of the 6 places for the girls: \( {}^{6}C_{2} \) = 15.
2. In those places girls stand in decreasing order (1 way); boys fill the rest in decreasing order (1 way).
3. Total = 15

Teacher's Note:
1. When the order within a group is fixed, divide by the factorial of the group size: 6!/(4! 2!) = 15.
2. This is the same as arranging BBBBGG as letters with repetition.

Answer: (d) 90

Explanation:
1. Without restriction the 6 people can be interviewed in 6! = 720 orders.
2. For each mother–child pair, exactly half the orders have the mother first: factor 1/2 for each of 3 pairs.
3. Required = 720/8 = 90

Teacher's Note:
1. A fixed relative order of 2 objects divides the total by 2!, applied independently to each pair.
2. Equivalently: choose positions for each pair \( {}^{6}C_{2} \) × \( {}^{4}C_{2} \) × \( {}^{2}C_{2} \) = 90.

Answer: (c) \( \frac{6!}{2!\,3!} \)

Explanation:
1. BANANA has 6 letters: A three times, N twice, B once.
2. Arrangements = \( \frac{6!}{3!\,2!} = 60 \)

Teacher's Note:
1. Divide by the factorial of the count of each repeated letter.
2. Multiplying by the factorials (option b) is a common slip; repetition reduces the count.

Answer: (c) \( \frac{11!}{(2!)^3} \)

Explanation:
1. MATHEMATICS has 11 letters with M, A and T each occurring twice.
2. Arrangements = \( \frac{11!}{2!\,2!\,2!} = \frac{11!}{(2!)^3} \) = 4989600

Teacher's Note:
1. The factorials of repeated letters are multiplied in the denominator, never added.
2. List the letter counts first: M2, A2, T2, H1, E1, I1, C1, S1.

Answer: (c) \( \frac{6!}{2!} \)

Explanation:
1. ARRANGE: A twice, R twice, N, G, E.
2. Treat RR as one unit: units RR, A, A, N, G, E (6 units, A twice).
3. Arrangements = \( \frac{6!}{2!} = 360 \) (RR inside the unit can be arranged in only 1 way)

Teacher's Note:
1. Identical letters inside a block cannot be rearranged among themselves, so no extra factor is needed.
2. Still divide by 2! for the two A's outside the block.

Answer: (b) 169194

Explanation:
1. Total arrangements (N twice): \( \frac{9!}{2!} = 181440 \).
2. With HIN: 7! = 5040; DUS: \( \frac{7!}{2!} = 2520 \); TAN: 7! = 5040. Sum = 12600.
3. Two patterns together: HIN & DUS: 5! = 120, DUS & TAN: 120, HIN & TAN: 5! = 120; sum = 360. All three: 3! = 6.
4. Required = 181440 − 12600 + 360 − 6 = 169194

Teacher's Note:
1. Use inclusion–exclusion: subtract single patterns, add back pairs, subtract the triple.
2. Each pattern block is treated as one letter; watch for repeated letters remaining outside the blocks.

Answer: (a) so that no two alike letters are together is 960; (b) so that the word starts with E but does not end with N is 900; (d) so that vowels occur in alphabetical order is 840

Explanation:
1. ENGINEER has E×3, N×2, G, I, R; total = \( \frac{8!}{3!\,2!} = 3360 \).
2. (a) Arrange N, N, G, I, R with the N's apart: 60 − 24 = 36, then put the 3 E's in 3 of the 6 gaps: 36 × 20 = 720. If the N's are adjacent (24 ways), one E must go between them and 2 E's in 2 of the other 5 gaps: 24 × 10 = 240. Total 960, true.
3. (b) Starting with E: \( \frac{7!}{2!\,2!} = 1260 \); starting with E and ending with N: \( \frac{6!}{2!} = 360 \); 1260 − 360 = 900, true.
4. (c) Ending with N: \( \frac{7!}{3!} = 840 \); neither = 3360 − 1260 − 840 + 360 = 1620 ≠ 2460, false.
5. (d) Vowels E, E, E, I in the fixed order EEEI: \( \frac{8!}{4!\,2!} = 840 \), true.

Teacher's Note:
1. For 'no two alike together' with two repeated letters, place the more numerous letter in gaps and handle the case where the other repeated pair is split by it.
2. A fixed order among k letters is handled by treating them as k identical letters.

Answer: (d) \( \frac{4!\,3!}{2} \)

Explanation:
1. Algebra has vowels A, E, A (in positions 1, 4, 7) and consonants L, G, B, R (positions 2, 3, 5, 6).
2. Vowels in their places: \( \frac{3!}{2!} = 3 \) ways; consonants in their places: 4! = 24 ways.
3. Total = \( \frac{4!\,3!}{2} = 72 \)

Teacher's Note:
1. 'Relative position unchanged' means vowel places stay vowel places and consonant places stay consonant places.
2. Divide by 2! because the two A's are identical.

Answer: (b) 1386000

Explanation:
1. Arrange the other 10 letters (A×4, B×3, D, E, F): \( \frac{10!}{4!\,3!} = 25200 \) ways.
2. These create 11 gaps; place the two identical C's in 2 different gaps: \( {}^{11}C_{2} \) = 55 ways.
3. Total = 25200 × 55 = 1386000

Teacher's Note:
1. Gap method: arrange the unrestricted letters first and drop the separated letters into distinct gaps.
2. Since the C's are identical, choose the gaps with \( {}^{11}C_{2} \), not \( {}^{11}P_{2} \).

Answer: (b) 240

Explanation:
1. First four digits (1, 2, 3, 3): \( \frac{4!}{2!} = 12 \) ways.
2. Fifth digit: 2 ways; sixth digit: 10 ways; the seventh is then fixed.
3. Total = 12 × 2 × 10 = 240

Teacher's Note:
1. A digit determined by another digit contributes a factor of 1.
2. Use the alike-objects formula for the first four digits because the two 3's are identical.

Answer: (b) \( \frac{(a+2b+3c+d)!}{a!\,(b!)^2\,(c!)^3} \)

Explanation:
1. Total number of books = a + 2b + 3c + d.
2. Identical groups: one group of a, two groups of b, three groups of c copies.
3. Arrangements = \( \frac{(a+2b+3c+d)!}{a!\,(b!)^2\,(c!)^3} \)

Teacher's Note:
1. Each set of identical copies contributes its own factorial to the denominator.
2. Count the total carefully: b copies of each of two titles means 2b copies in all.

Answer: (c) NAAIG

Explanation:
1. Letters: A, A, G, I, N. Words beginning with A: 4! = 24; with G: 4!/2! = 12; with I: 12. Total 48.
2. The 49th word is the first word beginning with N: NAAGI.
3. The 50th word is NAAIG.

Teacher's Note:
1. Count blocks of words by their first letter until you get close to the required position.
2. Within a block, list the next words in alphabetical order.

Answer: (c) 17^th

Explanation:
1. Letters: E, E, Q, U, U. Words starting with E: \( \frac{4!}{2!} = 12 \).
2. Starting with QE: \( \frac{3!}{2!} = 3 \). Starting with QUE: QUEEU comes before QUEUE: 1.
3. Rank = 12 + 3 + 1 + 1 = 17

Teacher's Note:
1. With repeated letters, each block size uses the alike-objects formula.
2. The final +1 accounts for the word itself.

Answer: (a) 2193

Explanation:
1. Letters in order: A, A, D, H, I, K, R. Words beginning with A: 6! = 720; D, H, I, K: 4 × 6!/2! = 1440.
2. R A _: RAA: 4! = 24; RAD A: 3! = 6; RADH A: 2! = 2; RADHI A K: 1 (RADHIAK).
3. Words before RADHIKA = 720 + 1440 + 24 + 6 + 2 + 1 = 2193

Teacher's Note:
1. When the first letter is A, the remaining six letters are all distinct, so 6! words; otherwise the two A's remain and we divide by 2!.
2. The question asks for words before RADHIKA, so do not add 1 at the end.

Answer: (c) 3720

Explanation:
1. Green: at least one of 5: 2⁵ − 1 = 31 ways.
2. Blue: at least one of 4: 2⁴ − 1 = 15 ways.
3. Red: any number (including none): 2³ = 8 ways.
4. Total = 31 × 15 × 8 = 3720

Teacher's Note:
1. Each distinct object is either chosen or not, giving 2ⁿ subsets; subtract 1 when at least one is required.
2. Red dyes have no condition, so the empty choice is allowed for them.

Answer: (b) 31248

Explanation:
1. Red: 2⁶ − 1 = 63; blue: 2⁵ − 1 = 31; green: 2⁴ = 16.
2. Total = 63 × 31 × 16 = 31248

Teacher's Note:
1. 'At least one' from a group of n distinct objects gives 2ⁿ − 1 choices.
2. Groups with no condition contribute 2ⁿ because choosing none of them is allowed.

Answer: (d) 63

Explanation:
1. Total ways of answering 3 questions = 4³ = 64.
2. Exactly one way gives all answers correct.
3. Ways to fail to get all correct = 64 − 1 = 63

Teacher's Note:
1. 'Fail to get all correct' is the complement of 'all correct'.
2. Each question has 4 independent choices, so multiply 4 × 4 × 4.

Answer: (b) 31

Explanation:
1. Each lamp is either on or off: 2⁵ = 32 ways.
2. The hall is not illuminated when all lamps are off: 1 way.
3. Required = 32 − 1 = 31

Teacher's Note:
1. This is the number of ways of selecting one or more objects from n distinct objects, 2ⁿ − 1.
2. Equivalently \( {}^{5}C_{1} \) + \( {}^{5}C_{2} \) + … + \( {}^{5}C_{5} \) = 31.

Answer: (a) 41

Explanation:
1. Let A, B serve together or not at all, and C, D refuse to serve together.
2. A and B both in: choose 3 more from the other 7, excluding choices with both C and D: \( {}^{7}C_{3} \) − \( {}^{5}C_{1} \) = 35 − 5 = 30.
3. A and B both out: choose 5 from the other 7, excluding those with both C and D: \( {}^{7}C_{5} \) − \( {}^{5}C_{3} \) = 21 − 10 = 11.
4. Total = 30 + 11 = 41

Teacher's Note:
1. Make cases on the 'together or not at all' pair first; the other condition is then handled by subtraction.
2. To exclude 'both C and D', fix both of them and count the selections of the rest.

Answer: (c) 8

Explanation:
1. Adjacent seats on one side of 3 seats: 2 pairs; 2 sides: 4 pairs; the two persons can sit in 2 orders: 8 ways.
2. The remaining 4 seats are filled from the other 5 people: \( {}^{5}P_{4} \) = 120 = 5! ways.
3. Total = 8 × 5!, so k = 8

Teacher's Note:
1. Count the positions of the restricted pair first, then arrange the others in the remaining seats.
2. Note that \( {}^{5}P_{4} \) = 5!/1! = 5!, which is why the answer is written as k(5!).

Answer: (d) 1440

Explanation:
1. Consonants D, P, M, L are arranged in 4! = 24 ways, creating 5 gaps.
2. Choose 2 different gaps for the two vowel pairs: \( {}^{5}C_{2} \) = 10.
3. The vowels E, E, A, A read from left to right can be arranged in \( \frac{4!}{2!\,2!} = 6 \) ways (first two form the left pair).
4. Total = 24 × 10 × 6 = 1440

Teacher's Note:
1. Putting the two vowel pairs in different gaps guarantees that consonants separate them.
2. Handle the identical E's and A's by arranging the vowel string with the alike-objects formula.

Answer: (a) 126

Explanation:
1. If India wins the series with Pakistan winning k matches (k = 0 to 4), India wins the last match and the first 4 + k matches contain exactly k Pakistan wins.
2. Ways = \( \sum_{k=0}^{4} {}^{4+k}C_{k} = 1 + 5 + 15 + 35 + 70 \) = 126

Teacher's Note:
1. The series ends as soon as India gets its 5th win, so the last match is always an India win.
2. Only the positions of Pakistan's wins among the earlier matches need to be chosen.

Answer: (c) 3¹⁰ − 1

Explanation:
1. For each question there are 3 options: leave it, attempt it, or attempt its alternative.
2. Total = 3¹⁰; subtract the case of leaving all questions.
3. Required = 3¹⁰ − 1

Teacher's Note:
1. List what can happen to a single question; the multiplication principle then gives the total.
2. 'One or more' means the empty selection must be removed.

Answer: (b) 64

Explanation:
1. All three of the same variety: 8 ways.
2. Exactly two of one variety and one of another: 8 × 7 = 56 ways.
3. Total = 8 + 56 = 64

Teacher's Note:
1. Ice creams of one variety are identical, so only the varieties chosen and their counts matter.
2. For the 'two of one kind' case, choose the repeated variety and then a different single variety.

Answer: (c) 1620

Explanation:
1. The number formed by the last two digits must be divisible by 4: 00, 04, 12, 20, 24, 32, 40, 44, 52, i.e. 9 ways.
2. First digit: 5 ways (1–5); second and third digits: 6 ways each.
3. Total = 5 × 6 × 6 × 9 = 1620

Teacher's Note:
1. Divisibility by 4 depends only on the last two digits.
2. List those two-digit endings carefully, remembering that the digits allowed are only 0–5.

Answer: (b) 36

Explanation:
1. Even digits: 0, 2, 4, 6, 8 (sum 20). Leaving out one digit, the sum must be a multiple of 3: leave out 2 or 8.
2. Digit sets {0, 4, 6, 8} and {0, 2, 4, 6}; each contains 0.
3. Numbers from each set = 4! − 3! = 18; total = 36

Teacher's Note:
1. Use the digit-sum test for 3 to choose the sets of digits first.
2. Subtract the arrangements starting with 0 for every set containing 0.

Answer: (c) 6ⁿ − 3ⁿ

Explanation:
1. Total outcomes = 6ⁿ.
2. Outcomes with no odd number (all even) = 3ⁿ.
3. Required = 6ⁿ − 3ⁿ

Teacher's Note:
1. 'At least one odd' is the complement of 'all even'.
2. Each die has 3 even faces, giving 3ⁿ all-even outcomes.

Answer: (c) 300

Explanation:
1. Consider 000 to 999 as three-digit strings: 1000 strings with 3000 digits in all.
2. Each digit 0–9 appears equally often: 3000/10 = 300 times; 1000 contains no 5.
3. So 5 is written 300 times.

Teacher's Note:
1. Counting 'how many times written' differs from 'how many numbers contain 5' (that is 271).
2. Padding with leading zeros makes every place symmetric without changing the count of 5's.

Answer: (b) 505

Explanation:
1. All 4-digit numbers with these digits (repetition allowed) = 5⁴ = 625.
2. Numbers with all digits different = \( {}^{5}P_{4} \) = 120.
3. Required = 625 − 120 = 505

Teacher's Note:
1. 'At least two identical' is the complement of 'all different'.
2. There is no 0, so every string of 4 digits is a valid number.

Answer: (c) mⁿ − m

Explanation:
1. Each prize can go to any of the m persons: mⁿ ways.
2. Not allowed: one person gets all n prizes: m ways.
3. Required = mⁿ − m

Teacher's Note:
1. Each prize 'chooses' a person, so the base count is mⁿ.
2. Receiving at most n − 1 prizes only rules out one person getting all n.

Answer: (d) 4 (7!)

Explanation:
1. Sum of 1 to 9 is 45; the two digits left out must have sum 9 (so that the remaining sum is a multiple of 9).
2. Pairs: (1, 8), (2, 7), (3, 6), (4, 5): 4 ways.
3. Each choice of 7 digits gives 7! numbers; total = 4 (7!)

Teacher's Note:
1. Divisibility by 9 depends only on the digit sum.
2. No zero is involved, so all 7! arrangements are valid.

Answer: (c) m (m − 1) (n − 5)^m−2

Explanation:
1. Choose the position of the maximum (n − 2): m ways; position of the minimum 2: (m − 1) ways.
2. Every other place takes a value strictly between 2 and n − 2, i.e. from 3 to n − 3: (n − 5) choices each.
3. Total = m (m − 1)(n − 5)^m−2

Teacher's Note:
1. Because max and min occur exactly once, the remaining entries must avoid both of them.
2. Positions of max and min are ordered (they are different numbers), giving m(m − 1).

Answer: (b) 375

Explanation:
1. 4-digit numbers starting with 1, 2 or 3: 3 × 5 × 5 × 5 = 375 (from 1000 to 3444).
2. Remove 1000 (not greater than 1000): 374.
3. Add 4000 (not greater than 4000 is allowed): 375

Teacher's Note:
1. Read the boundaries carefully: 'greater than 1000' excludes 1000, 'not greater than 4000' includes 4000.
2. Handle the boundary numbers separately after a clean count.

Answer: (b) 1800

Explanation:
1. Interpreting the condition as: no vowel lies between any two consonants, so all 5 consonants P, R, M, T, R stand together.
2. Consonant block: \( \frac{5!}{2!} = 60 \) arrangements; the block can occupy 5 positions among the 9 places (vowels on either side).
3. Vowels A, A, E, E in the remaining 4 places: \( \frac{4!}{2!\,2!} = 6 \).
4. Total = 60 × 5 × 6 = 1800

Teacher's Note:
1. If a vowel sat anywhere between two consonants, the consonants would not be consecutive, so the consonants must form a single block.
2. Repeated letters (R twice, A twice, E twice) must be divided out in each stage.

Answer: (c) 7

Explanation:
1. Real roots require q² ≥ 4p.
2. q = 1: none; q = 2: p = 1; q = 3: p = 1, 2; q = 4: p = 1, 2, 3, 4.
3. Total = 0 + 1 + 2 + 4 = 7

Teacher's Note:
1. Fix q and count the values of p that satisfy the discriminant condition.
2. Equal roots (discriminant zero) are real, so include equality.

Answer: (c) 20

Explanation:
1. Choose the 3 places for the girls: \( {}^{6}C_{3} \) = 20.
2. The order within each group is fixed by height.
3. Total = 20

Teacher's Note:
1. Fixed order inside a group means 6!/(3! 3!) arrangements.
2. Only the pattern of boy and girl positions matters.

Answer: (c) \( {}^{20}C_{10} \)

Explanation:
1. If each student solves r questions, at most \( {}^{20}C_{r} \) students can have different combinations.
2. \( {}^{20}C_{r} \) is greatest when r = 10.
3. Maximum = \( {}^{20}C_{10} \)

Teacher's Note:
1. \( {}^{n}C_{r} \) is maximum at r = n/2 when n is even.
2. The maximum over r is what the question asks for, since r is not given.

Answer: (d) None of these

Explanation:
1. Squares mod 5: 0 → 0, 1 → 1, 2 → 4, 3 → 4, 4 → 1.
2. Sum ≡ 0 (mod 5) needs both squares ≡ 0, or one ≡ 1 and the other ≡ 4.
3. Ordered pairs: (0, 0) and (1 or 4 with 2 or 3) in either order: 1 + 2 × 4 = 9 (unordered: 5).
4. Neither 9 nor 5 is among 13, 11, 17, so the answer is 'None of these'.

Teacher's Note:
1. Working with remainders on division by 5 keeps the counting small.
2. Whether the selection is ordered or not, check the result against each option before deciding.

Answer: (d) None of these

Explanation:
1. For a difference d (1 ≤ d ≤ 10), the pairs are (k, k + d) with k = 1 to 100 − d: (100 − d) pairs.
2. Total = \( \sum_{d=1}^{10} (100 - d) = 1000 - 55 = 945 \).
3. \( {}^{40}C_{2} \) = 780, \( {}^{70}C_{2} \) = 2415, \( {}^{100}C_{2} \) − \( {}^{99}C_{2} \) = 99; none equals 945.

Teacher's Note:
1. Classify the pairs by their difference; each difference gives an easy count.
2. Always evaluate the options numerically before choosing 'None of these'.

Answer: (b) 10

Explanation:
1. Number of trips = \( {}^{n}C_{3} \); trips including a particular child = \( {}^{n-1}C_{2} \).
2. \( {}^{n}C_{3} \) − \( {}^{n-1}C_{2} \) = \( {}^{n-1}C_{3} \) = 84.
3. \( {}^{9}C_{3} \) = 84, so n − 1 = 9 and n = 10.

Teacher's Note:
1. The trips without the particular child are exactly \( {}^{n-1}C_{3} \), which makes the equation simple.
2. Use \( {}^{n}C_3 = {}^{n-1}C_3 + {}^{n-1}C_2 \).

Answer: (d) 16

Explanation:
1. 4 members with no couple means exactly one member from each of the 4 couples.
2. Each couple gives husband or wife: 2 ways.
3. Total = 2⁴ = 16

Teacher's Note:
1. When the committee size equals the number of couples, avoiding couples forces one from each.
2. Check: \( {}^{8}C_{4} \) = 70 committees in all; complementary counting also gives 16.

Answer: (c) 286

Explanation:
1. Single letter: 26 names.
2. Letter followed by a digit: 26 × 10 = 260 names.
3. Total = 26 + 260 = 286

Teacher's Note:
1. The two forms are disjoint, so their counts are added.
2. Within the second form, letter and digit are chosen independently, so multiply.