CBSE Class 11 Mathematics Permutations and Combinations Competitive Exam MCQs Set 06

Answer: (a) 530

Explanation:
1. Letters in alphabetical order: C, C, E, I, K, R, T. Since C comes first, only words beginning with C come before CRICKET.
2. Words beginning CC, CE, CI, CK: 5! = 120 each, i.e., 480.
3. Words beginning CRC and CRE: 4! = 24 each, i.e., 48; then words beginning CRICE: 2! = 2.
4. Words before CRICKET = 480 + 48 + 2 = 530.

Teacher's Note:
1. Since C is the first letter alphabetically, only words beginning with C come before CRICKET.
2. Move letter by letter, counting words with a smaller letter at each position.

Answer: (d) None of these

Explanation:
1. Total routines = 7! = 5040.
2. A, B, C can be ordered in 3! = 6 ways, only one of which is A, B, C.
3. Required = \( \frac{7!}{3!} = 840 \), not among (a), (b), (c).

Teacher's Note:
1. When the relative order of k persons is fixed, divide the total by k!.
2. 210 = 7!/(3! 4!) would wrongly treat the others as identical.

Answer: (c) 91

Explanation:
1. Primes less than 30: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 (10 primes).
2. p/q with p ≠ q: \( {}^{10}P_{2} = 90 \) different values (distinct primes give distinct fractions).
3. p = q gives only the number 1. Total = 90 + 1 = 91.

Teacher's Note:
1. Fractions of distinct primes are already in lowest terms, so no two are equal.
2. All fractions p/p equal 1 and count only once.

Answer: (b) (4!)²

Explanation:
1. All 9 digits 1 to 9 are used; the middle digit must have 4 smaller and 4 larger digits, so it is 5.
2. Digits 1 to 4 fill the first four places in 4! ways; 6 to 9 fill the last four in 4! ways.
3. Total = (4!)².

Teacher's Note:
1. Nine different non-zero digits means every digit from 1 to 9 is used once.
2. Fixing the middle digit as 5 is the key observation.

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

Explanation:
1. \( \frac{{}^{n}C_{r}}{{}^{n}C_{r-1}} = \frac{n-r+1}{r} \), which is greater than 1 while r < (n + 1)/2.
2. So the values increase up to the middle and then decrease.
3. For even n the single middle term is r = n/2.

Teacher's Note:
1. The binomial coefficients rise to the middle and fall symmetrically.
2. For odd n there are two equal greatest values at r = (n − 1)/2 and (n + 1)/2.

Answer: (c) \( {}^{52}C_{4} \)

Explanation:
1. The sum is \( {}^{47}C_{4} + {}^{47}C_{3} + {}^{48}C_{3} + {}^{49}C_{3} + {}^{50}C_{3} + {}^{51}C_{3} \).
2. \( {}^{47}C_{4} + {}^{47}C_{3} = {}^{48}C_{4} \); then \( {}^{48}C_{4} + {}^{48}C_{3} = {}^{49}C_{4} \), and so on.
3. Finally \( {}^{51}C_{4} + {}^{51}C_{3} = {}^{52}C_{4} \).

Teacher's Note:
1. Write the sum in increasing order so that Pascal's rule ⁿCᵣ + ⁿCᵣ₋₁ = ⁿ⁺¹Cᵣ can be used repeatedly.
2. Each step raises the upper index by 1 while the lower index stays 4.

Answer: (b) 12

Explanation:
1. \( \frac{n(n-1)(n-2)}{6} = 220 \Rightarrow n(n-1)(n-2) = 1320 \)
2. 1320 = 12 × 11 × 10, so n = 12.

Teacher's Note:
1. Look for three consecutive integers whose product is 6 × 220.
2. Check: ¹²C₃ = 220.

Answer: (b) 10

Explanation:
1. \( \frac{{}^{n}C_{5}}{{}^{n-2}C_{5}} = \frac{n(n-1)}{(n-5)(n-6)} = \frac{9}{2} \)
2. 2n² − 2n = 9n² − 99n + 270 ⇒ 7n² − 97n + 270 = 0 ⇒ (n − 10)(7n − 27) = 0
3. n is a natural number, so n = 10.

Teacher's Note:
1. Divide the two symbols to cancel the common factorials.
2. Reject the fractional root.

Answer: (c) \( \frac{{}^{n}P_{r}}{r!} \)

Explanation:
1. Each selection of r objects can be arranged in r! ways.
2. So \( {}^{n}C_{r} \times r! = {}^{n}P_{r} \), giving \( {}^{n}C_{r} = \frac{{}^{n}P_{r}}{r!} \).

Teacher's Note:
1. Permutations = selections × arrangements of each selection.
2. This is the basic link between ⁿPᵣ and ⁿCᵣ.

Answer: (d) 4 or −3

Explanation:
1. ᵐC₂ = ᵐC₁₀ with 2 ≠ 10 ⇒ m = 2 + 10 = 12.
2. n² − n = 12 ⇒ n² − n − 12 = 0 ⇒ (n − 4)(n + 3) = 0.
3. n = 4 or −3 (both give n² − n = 12).

Teacher's Note:
1. Use ᵐCₓ = ᵐCᵧ ⇒ x = y or x + y = m.
2. Only n² − n needs to be a natural number, so n = −3 is also acceptable here.

Answer: (a) \( {}^{n+1}C_{r} \)

Explanation:
1. \( {}^{n}C_{r} + {}^{n}C_{r-1} = \frac{n!}{r!\,(n-r)!} + \frac{n!}{(r-1)!\,(n-r+1)!} \)
2. \( = \frac{n!}{r!\,(n-r+1)!}[(n-r+1) + r] = \frac{(n+1)!}{r!\,(n-r+1)!} = {}^{n+1}C_{r} \)

Teacher's Note:
1. This is Pascal's rule, the basis of Pascal's triangle.
2. Combinatorially: selections of r from n + 1 either exclude or include a fixed object.

Answer: (b) 3

Explanation:
1. r ≠ r + 2, so r + (r + 2) = 8 ⇒ r = 3.
2. \( {}^{3}C_{2} = 3 \).

Teacher's Note:
1. Equal combinations with different lower indices have lower indices adding up to n.
2. Then substitute r in the required expression.

Answer: (d) No value

Explanation:
1. ⁿC₁₆ needs n ≥ 16, and ²⁰Cₙ₊₂ needs n + 2 ≤ 20, i.e., n ≤ 18.
2. n = 16: ²⁰C₁₈ = 190 ≠ 1; n = 17: ²⁰C₁₉ = 20 ≠ 17; n = 18: ²⁰C₂₀ = 1 ≠ 153.
3. So no value of n satisfies the equation.

Teacher's Note:
1. First find the range of n for which both symbols are defined.
2. Then simply test the few possible values.

Answer: (a) \( {}^{16}C_{3} \)

Explanation:
1. \( {}^{15}C_{13} = {}^{15}C_{2} \)
2. \( {}^{15}C_{3} + {}^{15}C_{2} = {}^{16}C_{3} \) (Pascal's rule).

Teacher's Note:
1. Convert using ⁿCᵣ = ⁿCₙ₋ᵣ to get consecutive lower indices.
2. Then apply ⁿCᵣ + ⁿCᵣ₋₁ = ⁿ⁺¹Cᵣ.

Answer: (d) 5

Explanation:
1. r + (r + 2) = 10 ⇒ r = 4.
2. \( {}^{5}C_{4} = 5 \).

Teacher's Note:
1. Use x + y = n when ⁿCₓ = ⁿCᵧ and x ≠ y.
2. ⁵C₄ = ⁵C₁ = 5.

Answer: (b) 9

Explanation:
1. \( \frac{{}^{n}C_{r}}{{}^{n}C_{r-1}} = \frac{n-r+1}{r} = \frac{84}{36} = \frac{7}{3} \Rightarrow 3n - 10r = -3 \)
2. \( \frac{{}^{n}C_{r+1}}{{}^{n}C_{r}} = \frac{n-r}{r+1} = \frac{126}{84} = \frac{3}{2} \Rightarrow 2n - 5r = 3 \)
3. Solving, n = 9 and r = 3.

Teacher's Note:
1. Ratios of consecutive combinations give linear equations in n and r.
2. Check: ⁹C₂ = 36, ⁹C₃ = 84, ⁹C₄ = 126.

Answer: (a) n > 6

Explanation:
1. \( {}^{n}C_{3} + {}^{n}C_{4} = {}^{n+1}C_{4} \)
2. \( {}^{n+1}C_{4} > {}^{n+1}C_{3} \Rightarrow \frac{n+1-3}{4} > 1 \Rightarrow n - 2 > 4 \Rightarrow n > 6 \)

Teacher's Note:
1. Combine the left side with Pascal's rule first.
2. Then compare consecutive coefficients using the ratio (n − r + 1)/r.

Answer: (c) 6

Explanation:
1. Either r + 3 = 2r − 6 ⇒ r = 9, or (r + 3) + (2r − 6) = 15 ⇒ r = 6.
2. Of the given options, r = 6 (check: ¹⁵C₉ = ¹⁵C₆).

Teacher's Note:
1. Always consider both possibilities x = y and x + y = n.
2. r = 9 also works (¹⁵C₁₂ = ¹⁵C₁₂) but is not an option.

Answer: (d) \( \binom{n+2}{r} \)

Explanation:
1. Split: \( \left[\binom{n}{r} + \binom{n}{r-1}\right] + \left[\binom{n}{r-1} + \binom{n}{r-2}\right] \)
2. \( = \binom{n+1}{r} + \binom{n+1}{r-1} = \binom{n+2}{r} \)

Teacher's Note:
1. Split the middle term 2(ⁿCᵣ₋₁) into two parts to pair with its neighbours.
2. Pascal's rule is used twice.

Answer: (c) > 7

Explanation:
1. \( {}^{n-1}C_{3} + {}^{n-1}C_{4} = {}^{n}C_{4} \)
2. \( {}^{n}C_{4} > {}^{n}C_{3} \Rightarrow \frac{n-3}{4} > 1 \Rightarrow n > 7 \)

Teacher's Note:
1. Pascal's rule turns the left side into a single coefficient.
2. Use ⁿCᵣ/ⁿCᵣ₋₁ = (n − r + 1)/r to compare.

Answer: (d) \( \binom{n+1}{r+1} \)

Explanation:
1. \( \binom{n}{n-r} = \binom{n}{r} \)
2. \( \binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1} \)

Teacher's Note:
1. First use ⁿCₙ₋ᵣ = ⁿCᵣ.
2. Then Pascal's rule with consecutive lower indices r and r + 1.

Answer: (a) 12

Explanation:
1. r − 6 = 3r + 1 gives r = −3.5 (not possible).
2. So (r − 6) + (3r + 1) = 43 ⇒ 4r = 48 ⇒ r = 12.

Teacher's Note:
1. Reject the case that gives a non-integer value.
2. Check: ⁴³C₆ = ⁴³C₃₇.

Answer: (a) 11

Explanation:
1. C(n, 5) + C(n, 6) = C(n + 1, 6).
2. C(n + 1, 6) > C(n + 1, 5) ⇒ \( \frac{n+1-5}{6} > 1 \) ⇒ n > 10.
3. Least n = 11.

Teacher's Note:
1. Pascal's rule first, then compare consecutive coefficients.
2. For n = 10, C(11, 6) = C(11, 5), so strict inequality fails.

Answer: (b) 31

Explanation:
1. ⁵C₀ + ⁵C₁ + … + ⁵C₅ = 2⁵ = 32.
2. Removing ⁵C₀ = 1 gives 31.

Teacher's Note:
1. The sum of all ⁿCᵣ is 2ⁿ.
2. Equivalently 5 + 10 + 10 + 5 + 1 = 31.

Answer: (d) 231

Explanation:
1. 12 + 8 = n ⇒ n = 20.
2. C(22, 20) = C(22, 2) = 231.

Teacher's Note:
1. Use x + y = n when ⁿCₓ = ⁿCᵧ.
2. Reduce C(22, 20) to C(22, 2) for easy calculation.

Answer: (a) 816

Explanation:
1. r + (r − 10) = 20 ⇒ r = 15.
2. \( {}^{18}C_{15} = {}^{18}C_{3} = 816 \).

Teacher's Note:
1. The case r = r − 10 is impossible, so the indices must add up to 20.
2. ¹⁸C₃ = 18 × 17 × 16/6 = 816.

Answer: (a) 14 or 7

Explanation:
1. \( 2\,{}^{n}C_{5} = {}^{n}C_{4} + {}^{n}C_{6} \). Divide by \( {}^{n}C_{5} \): \( 2 = \frac{5}{n-4} + \frac{n-5}{6} \).
2. 12(n − 4) = 30 + (n − 5)(n − 4) ⇒ n² − 21n + 98 = 0.
3. (n − 7)(n − 14) = 0 ⇒ n = 7 or 14.

Teacher's Note:
1. Express both neighbours as multiples of the middle coefficient using the ratio formula.
2. Check: ⁷C₄, ⁷C₅, ⁷C₆ = 35, 21, 7 are in A.P.

Answer: (d) 165

Explanation:
1. The captain is always in the team, so 8 more players are needed.
2. They are chosen from the remaining 11 players: \( {}^{11}C_{8} = {}^{11}C_{3} = 165 \).

Teacher's Note:
1. A particular person always included reduces both n and r by 1.
2. ¹¹C₈ = ¹¹C₃ = 165.

Answer: (d) 4845

Explanation:
1. All 16 posts are alike, so we only select 16 applicants out of 20.
2. Ways = \( {}^{20}C_{16} = {}^{20}C_{4} = \frac{20 \times 19 \times 18 \times 17}{24} = 4845 \).

Teacher's Note:
1. All posts are identical (clerks), so this is a selection.
2. Use ⁿCᵣ = ⁿCₙ₋ᵣ to compute easily.

Answer: (b) 330

Explanation:
1. The particular lady is always included, so 4 more members are needed.
2. They are chosen from the remaining 11 persons: \( {}^{11}C_{4} = 330 \).

Teacher's Note:
1. Fixed inclusion: remove that person and reduce the size of the committee by one.
2. There is no other restriction on gentlemen or ladies.

Answer: (d) \( ({}^{5}C_{3} \times {}^{4}C_{2})(5)! \)

Explanation:
1. Select letters: \( {}^{5}C_{3} \times {}^{4}C_{2} \).
2. Arrange the 5 selected letters: 5!.
3. Words = \( {}^{5}C_{3} \times {}^{4}C_{2} \times 5! = 7200 \).

Teacher's Note:
1. Words need both selection and arrangement.
2. Select first, then multiply by the arrangements of the selected letters.

Answer: (c) 150

Explanation:
1. Female typist: 10 ways; male typist: 15 ways.
2. Total = 10 × 15 = 150.

Teacher's Note:
1. Both posts must be filled, so multiply.
2. Adding (25) would be the count for selecting only one typist.

Answer: (b) 12

Explanation:
1. Each handshake corresponds to a pair of persons: \( {}^{n}C_{2} = 66 \).
2. \( \frac{n(n-1)}{2} = 66 \Rightarrow n(n-1) = 132 = 12 \times 11 \Rightarrow n = 12 \).

Teacher's Note:
1. Each handshake corresponds to a pair of persons.
2. Find two consecutive integers with product 132.

Answer: (a) 6720

Explanation:
1. The guest's chair is fixed.
2. The other 5 persons are seated on the remaining 8 chairs: \( {}^{8}P_{5} = 6720 \).

Teacher's Note:
1. Fix the restricted person first.
2. Seating persons on chairs is an arrangement, so use ⁿPᵣ.

Answer: (b) \( {}^{10}C_{5} \times {}^{8}C_{4} \)

Explanation:
1. Red: \( {}^{10}C_{5} \); white: \( {}^{8}C_{4} \).
2. Total = \( {}^{10}C_{5} \times {}^{8}C_{4} \).

Teacher's Note:
1. Select from each colour separately and multiply.
2. Match each choice with the correct total of that colour.

Answer: (b) \( {}^{15}C_{2} \)

Explanation:
1. Each handshake is between a pair of persons.
2. Number of handshakes = \( {}^{15}C_{2} = \frac{15 \times 14}{2} = 105 \).

Teacher's Note:
1. A handshake between A and B is the same as between B and A.
2. So combinations, not permutations, are used.

Answer: (b) 9

Explanation:
1. If there are n teams, the number of matches = \( {}^{n}C_{2} = 36 \).
2. \( n(n-1) = 72 = 9 \times 8 \Rightarrow n = 9 \).

Teacher's Note:
1. Each match is decided by a pair of teams.
2. Solve n(n − 1) = 2 × 36.

Answer: (b) r = 4

Explanation:
1. \( \frac{{}^{2n}C_{3}}{{}^{n}C_{2}} = \frac{2n(2n-1)(2n-2)/6}{n(n-1)/2} = \frac{4(2n-1)}{3} = \frac{44}{3} \Rightarrow n = 6 \)
2. \( {}^{6}C_{r} = 15 \Rightarrow r = 2 \) or 4.
3. Among the options, r = 4.

Teacher's Note:
1. Cancel n(n − 1) to get a simple linear equation.
2. ⁶C₂ = ⁶C₄ = 15.

Answer: (c) {8, 9, 10}

Explanation:
1. \( \frac{{}^{10}C_{x}}{{}^{10}C_{x-1}} = \frac{11-x}{x} < \frac{1}{2} \Rightarrow 22 - 2x < x \Rightarrow x > \frac{22}{3} \)
2. Also 1 ≤ x ≤ 10. Hence x = 8, 9, 10.

Teacher's Note:
1. Use the ratio of consecutive combinations.
2. Remember x must keep both symbols defined, so x ≤ 10.

Answer: (d) \( 3\,{}^{m+1}C_{4} \)

Explanation:
1. \( \alpha = \frac{m(m-1)}{2} \); \( {}^{\alpha}C_{2} = \frac{\alpha(\alpha-1)}{2} = \frac{m(m-1)}{4}\cdot\frac{m^{2}-m-2}{2} \)
2. \( = \frac{(m+1)m(m-1)(m-2)}{8} = 3 \cdot \frac{(m+1)m(m-1)(m-2)}{24} = 3\,{}^{m+1}C_{4} \)

Teacher's Note:
1. Factor m² − m − 2 = (m − 2)(m + 1).
2. Check with m = 4: α = 6, ⁶C₂ = 15 = 3 × ⁵C₄.

Answer: (b) \( {}^{18}C_{4} \)

Explanation:
1. Sum = \( {}^{14}C_{4} + {}^{14}C_{3} + {}^{15}C_{3} + {}^{16}C_{3} + {}^{17}C_{3} \)
2. \( {}^{14}C_{4} + {}^{14}C_{3} = {}^{15}C_{4} \), \( {}^{15}C_{4} + {}^{15}C_{3} = {}^{16}C_{4} \), and so on up to \( {}^{18}C_{4} \).

Teacher's Note:
1. Arrange terms in increasing upper index before applying Pascal's rule.
2. Each step raises the upper index by 1.

Answer: (c) \( \frac{1}{2} n\,a_{n} \)

Explanation:
1. Let \( S = \sum_{r=0}^{n} \frac{r}{{}^{n}C_{r}} \). Replace r by n − r: \( S = \sum_{r=0}^{n} \frac{n-r}{{}^{n}C_{n-r}} = \sum_{r=0}^{n} \frac{n-r}{{}^{n}C_{r}} \).
2. Adding: \( 2S = \sum_{r=0}^{n} \frac{n}{{}^{n}C_{r}} = n\,a_{n} \)
3. \( S = \frac{1}{2} n\,a_{n} \).

Teacher's Note:
1. Writing the sum in reverse order and adding is a standard trick.
2. It works because ⁿCᵣ = ⁿCₙ₋ᵣ.