CBSE Class 11 Mathematics Permutations and Combinations Competitive Exam MCQs Set 05

Answer: (a) 420

Explanation:
1. The 7 digits are 1, 2, 2, 3, 3, 3, 4: 2 occurs twice and 3 thrice.
2. Numbers = \( \frac{7!}{2!\,3!} = \frac{5040}{12} = 420 \).

Teacher's Note:
1. All 7 digits are used in each number.
2. No 0 is present, so no arrangement needs to be removed.

Answer: (b) 108

Explanation:
1. Units digit: 1, 3 or 5 (3 ways).
2. Hundreds and tens digits: 6 ways each.
3. Total = 6 × 6 × 3 = 108.

Teacher's Note:
1. Fill the restricted units place first.
2. With repetition allowed, the other places keep all 6 choices.

Answer: (c) 60

Explanation:
1. Odd digits 3, 3, 5, 5 go into the 4 even places: \( \frac{4!}{2!\,2!} = 6 \) ways.
2. Even digits 2, 2, 8, 8, 8 go into the 5 odd places: \( \frac{5!}{2!\,3!} = 10 \) ways.
3. Total = 6 × 10 = 60.

Teacher's Note:
1. A nine-digit number has 4 even places and 5 odd places, matching the 4 odd digits exactly.
2. Arrange each group separately and multiply.

Answer: (c) 72

Explanation:
1. Units digit 2, 4 or 6: 3 ways.
2. Remaining 4 digits: 4! = 24 ways.
3. Total = 3 × 24 = 72.

Teacher's Note:
1. Fix the last digit for evenness, then arrange the rest.
2. Three of the five digits are even, so the answer is 3/5 of 5! = 120.

Answer: (a) 4⁶

Explanation:
1. Each ring can be worn on any of the 4 fingers.
2. Total = 4⁶ (ignoring the order of rings on a finger).

Teacher's Note:
1. Here the rings choose the fingers, so the base is the number of fingers.
2. A common error is 6⁴, which wrongly lets fingers choose rings.

Answer: (b) 3⁴

Explanation:
1. Each prize can go to any of the 3 students.
2. Total = 3⁴ = 81.

Teacher's Note:
1. The objects being distributed (prizes) form the exponent.
2. Since a student may get any number of prizes, each prize has full choice.

Answer: (b) 60

Explanation:
1. Total ways = 4³ = 64.
2. Ways with all letters in the same box = 4.
3. Required = 64 − 4 = 60.

Teacher's Note:
1. Use the complement: total minus the forbidden cases.
2. Each letter independently chooses one of 4 boxes.

Answer: (c) 5⁴

Explanation:
1. Each parcel can be registered at any of the 5 post-offices.
2. Total = 5⁴.

Teacher's Note:
1. Each parcel makes an independent choice among the post-offices.
2. The number of choices (5) is the base and the number of parcels (4) is the exponent.

Answer: (b) 810

Explanation:
1. Two digit numbers: 9 × 9 = 81.
2. Three digit numbers: 9 × 9 × 9 = 729.
3. Total = 81 + 729 = 810.

Teacher's Note:
1. Between 10 and 1000 means 2 digit and 3 digit numbers.
2. Without 0 in the digit set, every place has 9 choices.

Answer: (a) 69760

Explanation:
1. Total five-letter words with repetition allowed = 10⁵ = 100000.
2. Words with all letters different = \( {}^{10}P_{5} = 30240 \).
3. At least one repeated = 100000 − 30240 = 69760.

Teacher's Note:
1. 'At least one repeated' is the complement of 'all different'.
2. Words are arrangements, so use ¹⁰P₅ and not ¹⁰C₅.

Answer: (a) 20

Explanation:
1. Equal numbers of heads and tails among 6 coins means 3 heads and 3 tails.
2. Arrangements of HHHTTT = \( \frac{6!}{3!\,3!} = 20 \).

Teacher's Note:
1. Coins showing the same face are indistinguishable.
2. This also equals ⁶C₃, the number of ways to choose the positions of the heads.

Answer: (c) \( \frac{n(n^{r}-1)}{n-1} \)

Explanation:
1. Taking k things at a time with repetition gives nᵏ permutations.
2. Total = n + n² + … + nʳ, a G.P. with first term n and ratio n.
3. Sum = \( \frac{n(n^{r}-1)}{n-1} \).

Teacher's Note:
1. 'Not more than r' means add the cases 1, 2, …, r.
2. The sum of a G.P. a + ar + … (r terms) is a(rⁿ − 1)/(r − 1).

Answer: (c) 4096

Explanation:
1. Write every such integer with 4 digits using leading zeros (e.g., 25 as 0025).
2. Each of the 4 places can be any of the 8 digits: 8⁴ = 4096.

Teacher's Note:
1. Padding with leading zeros turns numbers of different lengths into 4-place strings.
2. If 0 itself is excluded, the count of positive integers is 4095.

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

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

Teacher's Note:
1. Each title with repeated copies contributes its own factorial to the denominator.
2. Two titles with b copies give (b!)² and three titles with c copies give (c!)³.

Answer: (b) \( \frac{(m+n)!}{m!\,n!} \)

Explanation:
1. A symmetrical arrangement is decided by one half, which contains m white and n red counters.
2. Arrangements of one half = \( \frac{(m+n)!}{m!\,n!} \); the other half is its mirror image.

Teacher's Note:
1. Symmetry means the second half is fixed once the first half is chosen.
2. So only m white and n red counters are actually arranged.

Answer: (d) 720

Explanation:
1. Units digit: 1, 3, 5 or 7 (4 ways).
2. Thousands digit: any non-zero digit (5 ways).
3. Hundreds and tens: 6 ways each.
4. Total = 5 × 6 × 6 × 4 = 720.

Teacher's Note:
1. Odd numbers must end in an odd digit; here there are four of them.
2. The first digit cannot be 0 even when repetition is allowed.

Answer: (c) \( \frac{12!}{5!\,3!\,2!} \times \frac{{}^{13}P_{3}}{3!} \)

Explanation:
1. Arrange the 12 letters other than C: \( \frac{12!}{5!\,3!\,2!} \) ways.
2. There are 13 gaps; choose 3 of them for the identical C's: \( {}^{13}C_{3} = \frac{{}^{13}P_{3}}{3!} \).
3. Total = \( \frac{12!}{5!\,3!\,2!} \times \frac{{}^{13}P_{3}}{3!} \).

Teacher's Note:
1. Use the gap method: arrange the others first, then place the C's in separate gaps.
2. The C's are identical, so choose gaps (combination) rather than arrange them.

Answer: (a) 4!

Explanation:
1. Consonants: M, X, M, M (4); vowels: A, I, U (3).
2. No two consonants together forces the pattern C V C V C V C.
3. Vowels: 3! = 6 ways; consonants: \( \frac{4!}{3!} = 4 \) ways. Total = 24 = 4!.

Teacher's Note:
1. With 4 consonants and 3 vowels, consonants must take the 1st, 3rd, 5th and 7th places.
2. Remember that M appears three times.

Answer: (c) 14400

Explanation:
1. Consonants T, R, N, G, L: 5! = 120 ways.
2. Vowels I, A, E go into 3 of the 6 gaps: \( {}^{6}P_{3} = 120 \).
3. Total = 120 × 120 = 14400.

Teacher's Note:
1. Gap method: arrange the unrestricted letters first.
2. n letters create n + 1 gaps including the two ends.

Answer: (a) 6!

Explanation:
1. C and Y are fixed at the ends.
2. The remaining 6 different letters are arranged in 6! ways.

Teacher's Note:
1. Fixed letters reduce the problem to arranging the rest.
2. The ends are specified, so they are not interchanged.

Answer: (b) 24

Explanation:
1. L is fixed in the 3rd place.
2. The other 4 letters fill 4 places in 4! = 24 ways.

Teacher's Note:
1. Place the fixed letter first, then arrange the rest.
2. All letters of DELHI are different.

Answer: (b) 900

Explanation:
1. Total arrangements = \( \frac{7!}{2!\,2!} = 1260 \) (A twice, R twice).
2. Both R together (treat RR as one unit): \( \frac{6!}{2!} = 360 \).
3. Required = 1260 − 360 = 900.

Teacher's Note:
1. Not together = Total − Together.
2. When the two identical R's are tied, they have only one internal arrangement.

Answer: (c) 240

Explanation:
1. Total = \( \frac{6!}{2!} = 360 \).
2. B and H together: \( \frac{5!}{2!} \times 2! = 120 \).
3. Required = 360 − 120 = 240.

Teacher's Note:
1. Tie B and H as one unit and remember they can swap inside it.
2. A appears twice in BHARAT.

Answer: (d) None of these

Explanation:
1. Vowels I, U, A, E form one block; remaining letters N, S, R, N, C.
2. 6 units with N twice: \( \frac{6!}{2!} = 360 \); vowels inside: 4! = 24.
3. Total = 360 × 24 = 8640, which is not among (a), (b), (c).

Teacher's Note:
1. INSURANCE contains N twice, so divide by 2!.
2. Without the 2! the answer would be 17280, a common slip.

Answer: (b) \( 7! \times {}^{8}P_{3} \)

Explanation:
1. Seat the 7 boys: 7! ways.
2. Girls in 3 of the 8 gaps: \( {}^{8}P_{3} \).
3. Total = \( 7! \times {}^{8}P_{3} \).

Teacher's Note:
1. Seven persons in a row create eight gaps.
2. Girls are different, so use ⁸P₃, not ⁸C₃.

Answer: (c) 5! × 6!

Explanation:
1. Boys: 5! ways; 6 gaps for 5 girls: \( {}^{6}P_{5} = 6! \).
2. Total = 5! × 6!.

Teacher's Note:
1. Gap method with 5 boys giving 6 gaps.
2. ⁶P₅ = 6!/1! = 6!.

Answer: (a) 40

Explanation:
1. Total = \( \frac{6!}{3!\,2!} = 60 \).
2. N's together: \( \frac{5!}{3!} = 20 \).
3. Required = 60 − 20 = 40.

Teacher's Note:
1. Use complement counting: Total − (N's together).
2. Alternatively, arrange B, A, A, A in 4 ways and choose 2 of 5 gaps: 4 × 10 = 40.

Answer: (a) 2880

Explanation:
1. Seat the boys: 5! = 120.
2. A girl between two boys can only be in the 4 inner gaps: \( {}^{4}P_{3} = 24 \).
3. Total = 120 × 24 = 2880.

Teacher's Note:
1. 'Between two boys' excludes the two end gaps.
2. So only 4 gaps are available for 3 girls.

Answer: (c) 144

Explanation:
1. Vowels A, I, E (3) and even places 2nd, 4th, 6th (3): 3! = 6 ways.
2. Consonants R, T, C, L in the other 4 places: 4! = 24.
3. Total = 144.

Teacher's Note:
1. Match the number of vowels with the number of even places.
2. Arrange both groups independently and multiply.

Answer: (c) \( {}^{4}P_{3} \)

Explanation:
1. The first student can get any of 4 grades, the second any of the remaining 3, the third any of the remaining 2.
2. Total = 4 × 3 × 2 = \( {}^{4}P_{3} = 24 \).

Teacher's Note:
1. Different students are distinguishable, so the order of assigned grades matters.
2. 4³ would allow two students the same grade.

Answer: (b) 420

Explanation:
1. Choose 2 men: \( {}^{7}C_{2} = 21 \).
2. Choose 2 women other than their wives: \( {}^{5}C_{2} = 10 \).
3. Pair them into two teams of a man and a woman: 2 ways. Total = 21 × 10 × 2 = 420.

Teacher's Note:
1. A mixed double has two teams, each of one man and one woman.
2. After choosing the men, their wives are excluded from the women's choice.

Answer: (d) 48

Explanation:
1. First digit 4: remaining 2, 5, 5, 7 give \( \frac{4!}{2!} = 12 \).
2. First digit 5: remaining 2, 4, 5, 7 give 4! = 24.
3. First digit 7: remaining 2, 4, 5, 5 give \( \frac{4!}{2!} = 12 \). Total = 48.

Teacher's Note:
1. Split by the first digit since that decides 'greater than 40000'.
2. Watch whether the repeated 5 is still among the remaining digits.

Answer: (b) (n − 1)!(n − 2)

Explanation:
1. Total = n!; together = 2 × (n − 1)!.
2. Not together = n! − 2(n − 1)! = (n − 1)![n − 2].

Teacher's Note:
1. Take (n − 1)! common to simplify.
2. The two specified books can swap within their unit, giving the factor 2.

Answer: (a) \( 5 \times {}^{8}P_{3} \)

Explanation:
1. Four digit numbers with first digit 5, 6, 7, 8 or 9: 5 ways.
2. Other 3 places from the remaining 8 digits: \( {}^{8}P_{3} \).
3. Total = \( 5 \times {}^{8}P_{3} \).

Teacher's Note:
1. Fill the first place according to the lower bound 5000.
2. Order matters in numbers, so use permutations for the remaining places.

Answer: (a) 9 × 9!

Explanation:
1. First digit: 9 ways (not 0).
2. Remaining 8 places from 9 digits: \( {}^{9}P_{8} = 9! \).
3. Total = 9 × 9!.

Teacher's Note:
1. Exclude 0 from the first place only.
2. Equivalently, 10! − 9! = 9 × 9!.

Answer: (c) 671

Explanation:
1. Total outcomes = 6⁴ = 1296.
2. No die shows 2: 5⁴ = 625.
3. At least one 2 = 1296 − 625 = 671.

Teacher's Note:
1. 'At least one' is best found by subtracting 'none' from the total.
2. Each die is distinct, so outcomes are ordered.

Answer: (a) 100

Explanation:
1. These are 3 digit numbers.
2. Hundreds digit: 5 ways (not 0); then 5 × 4 for the other places.
3. Total = 5 × 5 × 4 = 100.

Teacher's Note:
1. Between 99 and 1000 means exactly 3 digits.
2. 0 can be used in the tens or units place.

Answer: (c) 108

Explanation:
1. Total numbers = 4! = 24; each digit is in the units place 3! = 6 times.
2. Sum of unit digits = 6 × (3 + 4 + 5 + 6) = 108.

Teacher's Note:
1. Each digit occupies a given place equally often.
2. Multiplying by 1111 would give the sum of all the numbers.

Answer: (c) NAAIG

Explanation:
1. Words beginning with A: 4! = 24; with G: \( \frac{4!}{2!} = 12 \); with I: 12. Total 48.
2. The 49th word is the first starting with N: NAAGI.
3. The 50th word is NAAIG.

Teacher's Note:
1. Count blocks of words by their first letter in alphabetical order.
2. A appears twice, so blocks not starting with A are 4!/2! = 12.

Answer: (d) None of these

Explanation:
1. Women: \( {}^{4}P_{2} = 12 \) ways.
2. Men: 6 chairs remain, so \( {}^{6}P_{3} = 120 \) ways.
3. Total = \( {}^{4}P_{2} \times {}^{6}P_{3} = 1440 \), which is not in (a), (b), (c).

Teacher's Note:
1. After the women sit, 6 chairs remain for the men, not 4.
2. Chairs are numbered, so these are arrangements (permutations).

Answer: (b) 12

Explanation:
1. a = (x + 2)!, \( b = \frac{x!}{(x-11)!} \), c = (x − 11)!, so bc = x!.
2. (x + 2)! = 182 x! ⇒ (x + 2)(x + 1) = 182 = 14 × 13.
3. x + 2 = 14 ⇒ x = 12.

Teacher's Note:
1. The product bc collapses to x!, which simplifies the equation.
2. Recognise 182 as a product of consecutive integers.

Answer: (d) \( \frac{1}{2}(10!) \)

Explanation:
1. Total rankings = 10!.
2. In exactly half of them A₁ is above A₁₀ (by symmetry).
3. Required = \( \frac{1}{2}(10!) \).

Teacher's Note:
1. Swapping A₁ and A₁₀ pairs each ranking with another, so the two cases are equal.
2. Similarly, for three persons in a fixed order, divide by 3!.