CBSE Class 11 Mathematics Permutations and Combinations Competitive Exam MCQs Set 04

Answer: (d) 6 and 7

Explanation:
1. \( {}^{K+5}P_{K+1} = \frac{(K+5)!}{4!} \) and \( {}^{K+3}P_{K} = \frac{(K+3)!}{3!} \)
2. \( \frac{(K+5)(K+4)(K+3)!}{24} = \frac{11(K-1)}{2} \cdot \frac{(K+3)!}{6} \Rightarrow (K+5)(K+4) = 22(K-1) \)
3. K² − 13K + 42 = 0 ⇒ (K − 6)(K − 7) = 0 ⇒ K = 6 or 7.

Teacher's Note:
1. Convert both permutation symbols to factorials and cancel the common (K + 3)!.
2. The resulting quadratic gives two acceptable natural values.

Answer: (a) 25

Explanation:
1. Going to the town: 5 ways.
2. Returning (any of the 5 roads may be used again): 5 ways.
3. Total = 5 × 5 = 25.

Teacher's Note:
1. Nothing in the question forbids returning by the same road, so each journey has 5 choices.
2. If a different road had to be used for the return, the answer would be 5 × 4 = 20.

Answer: (d) 720

Explanation:
1. BHOPAL has 6 letters, all different.
2. Number of words = 6! = 720.

Teacher's Note:
1. A 'word' here means any arrangement of the letters, meaningful or not.
2. When all letters are distinct, the count is simply n!.

Answer: (d) \( {}^{4}P_{1} + {}^{4}P_{2} + {}^{4}P_{3} + {}^{4}P_{4} \)

Explanation:
1. Numbers can have 1, 2, 3 or 4 digits.
2. Counts: \( {}^{4}P_{1}, {}^{4}P_{2}, {}^{4}P_{3}, {}^{4}P_{4} \) respectively.
3. Total = \( {}^{4}P_{1} + {}^{4}P_{2} + {}^{4}P_{3} + {}^{4}P_{4} \) = 4 + 12 + 24 + 24 = 64.

Teacher's Note:
1. When the number of digits is not fixed, add the cases of each possible length.
2. There is no zero among the digits, so no case needs to be removed.

Answer: (d) 1024

Explanation:
1. Each question can be answered in 2 ways (true or false).
2. Total = 2¹⁰ = 1024.

Teacher's Note:
1. Independent two-way choices multiply, they do not add.
2. If at least one question must be answered, subtract 1 for the 'none answered' case only when blank answers are allowed.

Answer: (d) 8 × 7

Explanation:
1. Entering: 8 ways.
2. Leaving by a different gate: 7 ways.
3. Total = 8 × 7 = 56.

Teacher's Note:
1. Two stages done one after the other are combined by multiplication.
2. 'Different gate' means the entry gate cannot be used to exit.

Answer: (d) 24 ways

Explanation:
1. The 4 speakers P, Q, R, S are all different and all of them speak.
2. Orders of presentation = 4! = 4 × 3 × 2 × 1 = 24.

Teacher's Note:
1. Order of presentation is an arrangement of all 4 persons.
2. 256 = 4⁴ would wrongly allow the same speaker more than once.

Answer: (a) r!

Explanation:
1. Product of r consecutive numbers ending at n = n(n − 1)…(n − r + 1) = \( {}^{n}P_{r} \).
2. \( {}^{n}P_{r} = r! \times {}^{n}C_{r} \) and \( {}^{n}C_{r} \) is a natural number.
3. Hence the product is divisible by r!.

Teacher's Note:
1. Every product of r consecutive natural numbers is some ⁿPᵣ.
2. Example: 5 × 6 × 7 = 210 is divisible by 3! = 6.

Answer: (b) 60

Explanation:
1. The prizes are different, so we arrange 3 of the 5 competitors in the three prize positions.
2. First prize: 5 ways, second: 4 ways, third: 3 ways, i.e., \( {}^{5}P_{3} = 60 \).

Teacher's Note:
1. The prizes are different, so the order of the selected competitors matters.
2. One competitor cannot win two of these ranks, so repetition is not allowed.

Answer: (c) 720

Explanation:
1. The first passenger has 6 choices of seat, the next 5, and so on.
2. Total = 6 × 5 × 4 × 3 × 2 × 1 = 6! = 720.

Teacher's Note:
1. Seating all n persons in n seats is n!.
2. 36 = 6 × 6 is a common mistake that ignores the seats already taken.

Answer: (a) 1956

Explanation:
1. A signal uses r flags (r = 1 to 6) in order: \( {}^{6}P_{r} \) signals.
2. Total = 6 + 30 + 120 + 360 + 720 + 720 = 1956.

Teacher's Note:
1. A signal is an arrangement, so the order of flags matters.
2. Add the cases r = 1, 2, …, 6 because any number of flags may be used.

Answer: (a) \( \frac{(2n)!}{n!} \)

Explanation:
1. \( (2n)! = [1 \cdot 3 \cdot 5 \ldots (2n-1)] \times [2 \cdot 4 \cdot 6 \ldots 2n] \)
2. \( 2 \cdot 4 \cdot 6 \ldots 2n = 2^{n}\, n! \)
3. So \( 1 \cdot 3 \cdot 5 \ldots (2n-1) = \frac{(2n)!}{2^{n} n!} \) and \( 2^{n} \times \) this \( = \frac{(2n)!}{n!} \).

Teacher's Note:
1. Split (2n)! into its odd and even factors.
2. Taking 2 common from each of the n even factors leaves n!.

Answer: (b) 41

Explanation:
1. \( \frac{{}^{56}P_{r+6}}{{}^{54}P_{r+3}} = \frac{56!/(50-r)!}{54!/(51-r)!} = 56 \times 55 \times (51-r) \)
2. 3080(51 − r) = 30800 ⇒ 51 − r = 10 ⇒ r = 41.

Teacher's Note:
1. Use (51 − r)! = (51 − r)(50 − r)! to cancel the factorials.
2. 56 × 55 = 3080 is the key simplification.

Answer: (a) \( {}^{n-1}P_{r} + r\,{}^{n-1}P_{r-1} \)

Explanation:
1. Fix one particular object. Arrangements of r objects not containing it: ⁿ⁻¹Pᵣ.
2. Arrangements containing it: it takes one of r places and the other r − 1 places are filled in ⁿ⁻¹Pᵣ₋₁ ways, giving r · ⁿ⁻¹Pᵣ₋₁.
3. Hence \( {}^{n}P_{r} = {}^{n-1}P_{r} + r\,{}^{n-1}P_{r-1} \).

Teacher's Note:
1. Splitting into 'contains a fixed object' and 'does not contain it' proves many identities.
2. Check with n = 3, r = 2: ³P₂ = 6 and ²P₂ + 2 · ²P₁ = 2 + 4 = 6.

Answer: (c) 420

Explanation:
1. Units digit 0: other 3 places from 6 digits = 6 × 5 × 4 = 120.
2. Units digit 2, 4 or 6 (3 ways): thousands digit from 5 non-zero remaining digits, then 5 × 4 for the middle places = 3 × 5 × 5 × 4 = 300.
3. Total = 120 + 300 = 420.

Teacher's Note:
1. Treat the case 'last digit 0' separately, because 0 affects the choice of the first digit.
2. Always fill the most restricted places first.

Answer: (c) 90000

Explanation:
1. First digit: 9 ways (1 to 9); each of the other 4 digits: 10 ways.
2. Total = 9 × 10⁴ = 90000.

Teacher's Note:
1. The only restriction in forming numbers is that the first digit cannot be 0.
2. This also equals 99999 − 10000 + 1 = 90000.

Answer: (a) 156

Explanation:
1. Numbers less than 1000 have 1, 2 or 3 digits.
2. Count = \( {}^{6}P_{1} + {}^{6}P_{2} + {}^{6}P_{3} \) = 6 + 30 + 120 = 156.

Teacher's Note:
1. Split by number of digits when the length is not fixed.
2. No zero is present, so every arrangement is a valid number.

Answer: (a) 224

Explanation:
1. Units digit must be 2, 4, 6 or 8: 4 ways.
2. Remaining two places from 8 digits: 8 × 7 = 56.
3. Total = 4 × 56 = 224.

Teacher's Note:
1. Fill the units place first because it decides whether the number is even.
2. There is no 0 among the digits, so the hundreds place has no extra restriction.

Answer: (a) 216

Explanation:
1. Sum 0 + 1 + 2 + 3 + 4 + 5 = 15; the five digits used must have sum divisible by 3, so the omitted digit is 0 or 3.
2. Omit 0: digits 1, 2, 3, 4, 5 give 5! = 120 numbers.
3. Omit 3: digits 0, 1, 2, 4, 5 give 5! − 4! = 96 numbers.
4. Total = 120 + 96 = 216.

Teacher's Note:
1. A number is divisible by 3 when the sum of its digits is divisible by 3.
2. When 0 is among the digits, subtract the arrangements starting with 0.

Answer: (b) 86400

Explanation:
1. The 4 small cages must get 4 of the 5 animals that can enter them: \( {}^{5}P_{4} = 120 \) ways.
2. The remaining 6 animals go into the remaining 6 cages: 6! = 720 ways.
3. Total = 120 × 720 = 86400.

Teacher's Note:
1. Fill the restricted cages first with only the eligible animals.
2. After that, all remaining cages and animals are unrestricted.

Answer: (b) 12

Explanation:
1. The number is of the form 3 _ _ 5.
2. The two middle places are filled from the remaining 4 digits: \( {}^{4}P_{2} = 12 \).

Teacher's Note:
1. Between 3000 and 4000 fixes the first digit as 3.
2. Divisible by 5 with these digits means the last digit is 5.

Answer: (c) 10

Explanation:
1. Ending in 0: remaining 3 digits in 3! = 6 ways.
2. Ending in 2: first digit 1 or 3 (2 ways), then 2 ways for the rest = 4.
3. Total = 6 + 4 = 10.

Teacher's Note:
1. Split the even case by the last digit, keeping 0 separate.
2. When the last digit is 2, 0 cannot be the first digit.

Answer: (b) 4500000

Explanation:
1. Total seven digit numbers = 9 × 10⁶.
2. Fix the first six digits; of the 10 choices for the last digit, exactly 5 make the sum even.
3. Required = 9 × 10⁵ × 5 = 4500000.

Teacher's Note:
1. For any choice of the first six digits, half of the units digits give an even total.
2. So exactly half of all seven digit numbers qualify.

Answer: (a) 133320

Explanation:
1. There are 4! = 24 numbers; each digit appears 6 times in each place.
2. Sum of digits in each place = 6 × (2 + 4 + 6 + 8) = 120.
3. Total = 120 × (1000 + 100 + 10 + 1) = 120 × 1111 = 133320.

Teacher's Note:
1. Each digit occupies each place (n − 1)! times.
2. Multiply the column sum by 1111 to combine all four places.

Answer: (c) 84

Explanation:
1. First digit 3, 4 or 5: 3 × 4! = 72.
2. First digit 2, second digit 4 or 5: 2 × 3! = 12.
3. Total = 72 + 12 = 84.

Teacher's Note:
1. All such numbers are 5 digit numbers, so compare digit by digit from the left.
2. Numbers starting 21 or 23 are less than 24000 and are not counted.

Answer: (b) 12

Explanation:
1. The number must end in 5.
2. 3 digit numbers: \( {}^{3}P_{2} = 6 \); 4 digit numbers: 3! = 6.
3. Total = 12.

Teacher's Note:
1. Divisible by 5 with these digits forces the last digit to be 5.
2. Greater than 100 means 3 or 4 digits.

Answer: (a) 106656

Explanation:
1. Numbers greater than 1000 are the 4 digit numbers: 4! = 24.
2. Each digit appears 6 times in each place: column sum = 6 × 16 = 96.
3. Total = 96 × 1111 = 106656.

Teacher's Note:
1. With 4 distinct digits, only 4 digit numbers exceed 1000.
2. The same column-sum method works for any set of distinct digits without 0.

Answer: (c) \( \frac{12!}{(3!)^4} \)

Explanation:
1. Total books = 12, with 4 groups of 3 identical copies.
2. Arrangements = \( \frac{12!}{3!\,3!\,3!\,3!} = \frac{12!}{(3!)^4} \).

Teacher's Note:
1. Copies of the same book are identical, so divide by 3! for each title.
2. This is the formula n!/(p₁! p₂! … ) for repeated objects.

Answer: (c) 5! 4! 3! 2!

Explanation:
1. Treat each subject as one block: 3 blocks can be arranged in 3! ways.
2. Within blocks: 5!, 4! and 2! ways.
3. Total = 5! 4! 3! 2!.

Teacher's Note:
1. Tie each subject's books together, arrange the blocks, then arrange inside each block.
2. The books are different, so internal arrangements are counted.

Answer: (d) 1630

Explanation:
1. 1 digit: 5; 2 digits: 5 × 5 = 25; 3 digits: 5 × 5 × 4 = 100;
2. 4 digits: 5 × 5 × 4 × 3 = 300; 5 digits: 5 × 5 × 4 × 3 × 2 = 600; 6 digits: 5 × 5! = 600.
3. Total = 5 + 25 + 100 + 300 + 600 + 600 = 1630.

Teacher's Note:
1. For each length, the first digit has 5 choices (not 0), and the rest are filled from what remains.
2. 0 alone is not a positive integer, so the 1 digit count is 5.

Answer: (c) 3n² − 2n

Explanation:
1. Total triplets (repetition allowed) = n³.
2. Triplets with all coordinates different = \( {}^{n}P_{3} \).
3. At least two equal = \( n^{3} - {}^{n}P_{3} = n^{3} - n(n-1)(n-2) = 3n^{2} - 2n \).

Teacher's Note:
1. 'At least two equal' is the complement of 'all different'.
2. Counting the complement avoids handling the cases of exactly two equal and all three equal separately.

Answer: (d) 11

Explanation:
1. Fractions p/q with p < q: q = 2: 1; q = 3: 2; q = 4: 3; q = 5: 4; q = 6: 5; total 15.
2. Repeated values: 2/4 = 3/6 = 1/2, 2/6 = 1/3, 4/6 = 2/3 (4 repeats).
3. Distinct rationals = 15 − 4 = 11.

Teacher's Note:
1. Count all fractions first, then remove those equal to a fraction already counted.
2. Equal fractions are the same rational number.

Answer: (d) None of these

Explanation:
1. If the middle digit is n, the other 4 places are filled from the n smaller digits 0, 1, …, n − 1 in \( {}^{n}P_{4} \) ways; those with 0 in the first place number \( {}^{n-1}P_{3} \).
2. Required = \( \sum_{n=4}^{9} {}^{n}P_{4} - \sum_{n=4}^{9} {}^{n-1}P_{3} \) = 6048 − 756.
3. = 5292, which is not given in options (a), (b), (c).

Teacher's Note:
1. Option (a) forgets to remove numbers whose first digit is 0.
2. Always check whether 0 can land in the leading place.

Answer: (b) 243

Explanation:
1. Each match has 3 possible results.
2. Number of different forecasts for 5 matches = 3⁵ = 243.
3. To be sure that one person is fully correct, all 243 forecasts must be made, so n = 243.

Teacher's Note:
1. Since no two people make the same forecast, every possible forecast must be covered to guarantee a correct one.
2. 3⁴ = 81 would cover only 4 matches.

Answer: (b) \( \frac{9!}{2!\,4!\,3!} \)

Explanation:
1. x²y⁴z³ = x·x·y·y·y·y·z·z·z: 9 letters with x twice, y four times and z three times.
2. Permutations = \( \frac{9!}{2!\,4!\,3!} \).

Teacher's Note:
1. Write the powers out as repeated letters before counting.
2. Divide by the factorial of each repetition count.

Answer: (b) 60

Explanation:
1. Digits: 3, 4, 7, 5, 5 (5 repeated twice).
2. Numbers = \( \frac{5!}{2!} = 60 \).

Teacher's Note:
1. Identical digits are handled exactly like identical letters.
2. No zero is involved, so every arrangement is a valid 5 digit number.

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

Explanation:
1. SERIES: S twice, E twice, R, I once each (6 letters).
2. Arrangements = \( \frac{6!}{2!\,2!} = 180 \).

Teacher's Note:
1. List the repeated letters carefully: S and E both appear twice.
2. Distinct letters contribute 1! = 1 in the denominator.

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

Explanation:
1. MATHEMATICS has 11 letters: M, A, T twice each; H, E, I, C, S once.
2. Words = \( \frac{11!}{2!\,2!\,2!} \).

Teacher's Note:
1. Three letters repeat, so three factors of 2! appear in the denominator.
2. Strictly, 'rearranging' excludes the original word, but such options ignore that one word.

Answer: (a) 16800

Explanation:
1. Vowels: I, E, E, E, E (5); consonants: N, N, N, D, D, P, C (7).
2. Treat the vowels as one block: 8 units with N thrice and D twice: \( \frac{8!}{3!\,2!} = 3360 \).
3. Vowels inside the block: \( \frac{5!}{4!} = 5 \). Total = 3360 × 5 = 16800.

Teacher's Note:
1. Tie the vowels together, arrange the units, then arrange the vowels within the block.
2. Remember the repetitions both outside (N, D) and inside (E) the block.

Answer: (a) 1260

Explanation:
1. There are 10 balls in all: 5 red (alike), 4 blue (alike) and 1 green.
2. Arrangements = \( \frac{10!}{5!\,4!\,1!} = 1260 \).

Teacher's Note:
1. Balls of the same colour are treated as identical.
2. This is the same formula used for words with repeated letters.

Answer: (c) 5³

Explanation:
1. Each journey can be made by any of the 5 conveyances.
2. Total = 5 × 5 × 5 = 5³.

Teacher's Note:
1. The same conveyance may be used again, so each journey has 5 choices.
2. Count choices per stage and multiply: (choices)^(number of stages).

Answer: (a) 60

Explanation:
1. BANANA has 6 letters: A three times, N twice, B once.
2. Permutations = \( \frac{6!}{3!\,2!} = \frac{720}{12} = 60 \).

Teacher's Note:
1. Count each repeated letter before applying n!/(p! q!).
2. 720 = 6! would treat all letters as different.