CBSE Class 11 Mathematics Permutations and Combinations Competitive Exam MCQs Set 03

Answer: (b) \( \frac{(n-1)(n-2)}{2} \)

Explanation:
1. Any two of the concurrent lines meet at the common point, so a triangle needs the new line as one side.
2. The new line is parallel to one of the n lines, so it cuts only the other n − 1 lines.
3. Triangles = \( {}^{n-1}C_{2} \) = \( \frac{(n-1)(n-2)}{2} \)

Teacher's Note:
1. Three concurrent lines do not form a triangle, and parallel lines do not meet.
2. So choose 2 of the n − 1 lines that the extra line actually cuts.

Answer: (b) \( {}^{a+b+c}C_{3} \) − (\( {}^{a}C_{3} \) + \( {}^{b}C_{3} \) + \( {}^{c}C_{3} \))

Explanation:
1. Total selections of 3 points = \( {}^{a+b+c}C_{3} \).
2. Selections from the same side are collinear: \( {}^{a}C_{3} \) + \( {}^{b}C_{3} \) + \( {}^{c}C_{3} \).
3. Triangles = \( {}^{a+b+c}C_{3} \) − (\( {}^{a}C_{3} \) + \( {}^{b}C_{3} \) + \( {}^{c}C_{3} \))

Teacher's Note:
1. Points on one side of the triangle are collinear, so they cannot form a triangle.
2. The vertices A, B, C are not included since only interior points are used.

Answer: (b) 1023

Explanation:
1. Each lamp is on or off: 2¹⁰ = 1024 ways.
2. Exclude the case when all lamps are off.
3. Required = 1024 − 1 = 1023

Teacher's Note:
1. Selecting one or more of n distinct objects can be done in 2ⁿ − 1 ways.
2. The hall is illuminated as soon as at least one lamp is on.

Answer: (b) 30

Explanation:
1. INTEGER: 7 letters with E twice. Total words = \( \frac{7!}{2!} = 2520 \).
2. I and N together: \( \frac{6! \times 2!}{2!} = 720 \), so m₁ = 2520 − 720 = 1800.
3. m₂: I first and R last, middle N, T, E, G, E: \( \frac{5!}{2!} = 60 \).
4. m₁/m₂ = 1800/60 = 30

Teacher's Note:
1. 'Never together' is found as Total − Together.
2. Fixing letters at the ends reduces the problem to arranging the remaining letters.

Answer: (a) 72

Explanation:
1. Letters: S×4, A×3, I×2, N×2, T, O (6 kinds).
2. All four alike: SSSS, 1. Three alike + one different: 2 × 5 = 10. Two pairs: \( {}^{4}C_{2} \) = 6.
3. One pair + two different: 4 × \( {}^{5}C_{2} \) = 40. All different: \( {}^{6}C_{4} \) = 15.
4. Total = 1 + 10 + 6 + 40 + 15 = 72

Teacher's Note:
1. Classify selections by the pattern of repeated letters, and use only letters available in enough copies.
2. In a selection the order does not matter, so no arrangement factor is needed.

Answer: (a) 236

Explanation:
1. Letters: I, I, R, S, T, U. Words beginning with I: 5! = 120; with R: \( \frac{5!}{2!} = 60 \). Total 180.
2. SI: 4! = 24; SR: \( \frac{4!}{2!} = 12 \); ST: 12; total 228. SUI: 3! = 6; total 234.
3. SUR: SURIIT comes first (235), then SURITI.
4. Rank = 236

Teacher's Note:
1. Recompute the repeated-letter factor each time one I is used up.
2. Go letter by letter, counting only smaller unused letters at each position.

Answer: (a) \( \sum_{k=1}^{n-1} k \)

Explanation:
1. Each pair of lines meets in exactly one point and no point is shared by three lines.
2. Number of points = \( {}^{n}C_2 = \frac{n(n-1)}{2} = 1 + 2 + \ldots + (n-1) = \sum_{k=1}^{n-1} k \)

Teacher's Note:
1. No two parallel means every pair meets; no three concurrent means the points are all different.
2. Recognise \( \frac{n(n-1)}{2} \) as the sum of the first n − 1 natural numbers.

Answer: (a) 6

Explanation:
1. With b bananas (b ≥ 1), the rest 3 − b fruits are apples and oranges (at most 2 apples).
2. b = 1: (apples, oranges) = (0, 2), (1, 1), (2, 0): 3 ways. b = 2: (0, 1), (1, 0): 2 ways. b = 3: 1 way.
3. Total = 3 + 2 + 1 = 6

Teacher's Note:
1. Since fruits of a kind are alike, a selection is just the count of each kind.
2. Listing the cases by the number of bananas avoids double counting.

Answer: (a) 2454

Explanation:
1. EXAMINATION: A×2, I×2, N×2, E, X, M, T, O (8 kinds).
2. All different: \( {}^{8}P_{4} \) = 1680.
3. One pair + two different: 3 × \( {}^{7}C_{2} \) × \( \frac{4!}{2!} \) = 3 × 21 × 12 = 756.
4. Two pairs: \( {}^{3}C_{2} \) × \( \frac{4!}{2!\,2!} \) = 18. Total = 1680 + 756 + 18 = 2454

Teacher's Note:
1. First choose the letters by pattern, then arrange them, dividing by factorials of repeated letters.
2. Only A, I, N are available twice, so only they can form pairs.

Answer: (b) 292

Explanation:
1. Choose the middle square and then 2 of its neighbouring squares (sharing a side).
2. 4 corner squares have 2 neighbours: 1 way each = 4; 24 edge (non-corner) squares have 3 neighbours: \( {}^{3}C_{2} \) = 3 ways each = 72.
3. 36 inner squares have 4 neighbours: \( {}^{4}C_{2} \) = 6 ways each = 216.
4. Total = 4 + 72 + 216 = 292

Teacher's Note:
1. Classify the squares of the board by how many neighbours they have.
2. Each valid set has exactly one square touching both others, so nothing is counted twice.

Answer: (a) \( \frac{10!}{2} \); (b) 8! × \( {}^{10}C_{2} \)

Explanation:
1. Total rankings = 10!; in exactly half of them A₁ is ahead of A₂: \( \frac{10!}{2} \).
2. Alternatively, choose 2 ranks for A₁, A₂ (\( {}^{10}C_{2} \), the better one to A₁) and arrange the other 8: 8! × \( {}^{10}C_{2} \).
3. Both equal 1814400, so (a) and (b) are correct.

Teacher's Note:
1. Fixing the relative order of two objects halves the number of arrangements.
2. Counting the same thing in two ways is a good check for multi-correct questions.

Answer: (c) 13

Explanation:
1. In an AP, a + c = 2b, so 3b = 21 and b = 7.
2. Then a + c = 14 with a, c natural numbers: a = 1, 2, …, 13.
3. Number of triplets = 13

Teacher's Note:
1. The middle term of three terms in AP is their average.
2. Natural numbers start from 1, so a = 0 and a = 14 are not allowed.

Answer: (d) None of these

Explanation:
1. Choose the 2 elements that lie in both P and Q: \( {}^{n}C_{2} \) ways.
2. Each remaining element is in P only, in Q only, or in neither: 3 ways.
3. Number of ways = \( {}^{n}C_{2} \) · 3ⁿ⁻², which matches none of (a), (b), (c).

Teacher's Note:
1. Decide the status of each element independently; this is the key idea for counting pairs of subsets.
2. An element of neither type 'both' must avoid being in P and Q together, leaving 3 choices.

Answer: (b) 26 × 6!

Explanation:
1. The boxes are in 3 rows of 4, 2 and 2. Choose 6 of the 8 boxes: \( {}^{8}C_{6} \) = 28.
2. Leaving 2 boxes empty empties a row only when both boxes of row 2 or both of row 3 are left: 2 ways. Valid choices = 26.
3. Place the 6 distinct letters in the chosen boxes: 6! ways. Total = 26 × 6!

Teacher's Note:
1. Count the choice of empty boxes first, since exactly 2 boxes remain empty.
2. Row 1 has 4 boxes and can never be completely empty when only 2 boxes are unused.

Answer: (b) 53

Explanation:
1. Group stage: 2 × \( {}^{6}C_{2} \) = 30 matches.
2. Next round (6 teams): \( {}^{6}C_{2} \) = 15; semifinal round (4 teams): \( {}^{4}C_{2} \) = 6.
3. Final, best of three: at least 2 matches.
4. Minimum total = 30 + 15 + 6 + 2 = 53

Teacher's Note:
1. A round-robin among k teams needs \( {}^{k}C_{2} \) matches.
2. 'Minimum' refers to the final ending in 2 matches when one team wins both.

Answer: (c) n³

Explanation:
1. Each line contains n + 1 of the points (including O).
2. Triangles = \( {}^{2n+1}C_3 - 2\,{}^{n+1}C_3 \).
3. Directly: O with one A and one B: n²; two A's and one B or one A and two B's: \( 2n\,{}^{n}C_2 = n^2(n-1) \).
4. Total = n² + n³ − n² = n³

Teacher's Note:
1. Point O lies on both lines, so it must be counted with each line when removing collinear triples.
2. Check with n = 1: points O, A₁, B₁ give 1 triangle = 1³.

Answer: (d) 342

Explanation:
1. Choose 3 of the 4 sides and one point on each.
2. 3·4·5 + 3·4·6 + 3·5·6 + 4·5·6 = 60 + 72 + 90 + 120
3. = 342

Teacher's Note:
1. Vertices on different sides are never collinear, so every such choice gives a triangle.
2. Sum the products over all 4 ways of choosing 3 sides.

Answer: (b) 791

Explanation:
1. A 9 × 6 grid has 10 vertical and 7 horizontal lines: rectangles = \( {}^{10}C_{2} \) × \( {}^{7}C_{2} \) = 45 × 21 = 945.
2. Squares: 9·6 + 8·5 + 7·4 + 6·3 + 5·2 + 4·1 = 54 + 40 + 28 + 18 + 10 + 4 = 154.
3. Rectangles that are not squares = 945 − 154 = 791

Teacher's Note:
1. A rectangle is determined by 2 vertical and 2 horizontal grid lines.
2. k × k squares in a 9 × 6 grid number (10 − k)(7 − k) for k = 1 to 6.

Answer: (d) 36 (55³)

Explanation:
1. Unit's, ten's and hundred's places: digit pairs (x, y) with x + y ≤ 9: 10 + 9 + … + 1 = 55 each.
2. Thousand's place: both digits from 1 to 9 with x + y ≤ 9: 8 + 7 + … + 1 = 36.
3. Total = 36 × 55³

Teacher's Note:
1. 'No carrying' means the digits in each column add up to at most 9.
2. The thousand's digits must be non-zero for both numbers to be 4-digit numbers.

Answer: (b) 156

Explanation:
1. Every 3 points determine one circle: \( {}^{11}C_{3} \) = 165.
2. The \( {}^{5}C_{3} \) = 10 triples from the 5 concyclic points all give the same circle.
3. Circles = 165 − 10 + 1 = 156

Teacher's Note:
1. This is the circle version of 'lines through n points with m collinear'.
2. For the maximum, assume no three of the points are collinear.

Answer: (d) \( {}^{98}C_{3} \)

Explanation:
1. Selecting 3 books leaves 97 books, which create 98 gaps (including the ends).
2. No two selected books are neighbours exactly when they occupy 3 different gaps: \( {}^{98}C_{3} \) ways.

Teacher's Note:
1. Gap method: arrange the unselected objects and choose separate gaps for the selected ones.
2. In general, k non-adjacent objects from n in a row can be chosen in \( {}^{n-k+1}C_{k} \) ways.

Answer: (a) 50

Explanation:
1. \( A_n = {}^{n}C_{25} \times 25! \) (choose 25 students for room A and arrange them).
2. \( A_n - A_{n-1} = 25!\,({}^{n}C_{25} - {}^{n-1}C_{25}) = 25!\,{}^{n-1}C_{24} \).
3. \( {}^{n-1}C_{24} = {}^{49}C_{25} = {}^{49}C_{24} \) ⇒ n − 1 = 49 ⇒ n = 50

Teacher's Note:
1. Use \( {}^{n}C_{r} - {}^{n-1}C_{r} = {}^{n-1}C_{r-1} \) to simplify the difference.
2. Remember \( {}^{49}C_{25} = {}^{49}C_{24} \) when comparing.

Answer: (c) 106656

Explanation:
1. Numbers greater than 1000 must be 4-digit: 4! = 24 numbers.
2. Each digit appears 3! = 6 times in each place.
3. Sum = 6 × (1 + 3 + 5 + 7) × 1111 = 6 × 16 × 1111 = 106656

Teacher's Note:
1. The place-value method: (times each digit appears) × (sum of digits) × 1111.
2. Only 4 digits are available, so no 5-digit numbers exist.

Answer: (a) 6660

Explanation:
1. There are \( {}^{4}P_{3} \) = 24 numbers.
2. Each digit appears in each place \( {}^{3}P_{2} \) = 6 times.
3. Sum = 6 × (1 + 2 + 3 + 4) × 111 = 6 × 10 × 111 = 6660

Teacher's Note:
1. When a digit is fixed in one place, the other two places are filled from the remaining 3 digits in \( {}^{3}P_{2} \) ways.
2. Multiply by 111 to combine the unit's, ten's and hundred's places.

Answer: (d) \( \frac{(m+n-2)!}{(m-1)!\,(n-1)!} \)

Explanation:
1. A shortest path uses (m − 1) blocks eastward and (n − 1) blocks southward.
2. The number of paths is the number of arrangements of these (m + n − 2) steps: \( \frac{(m+n-2)!}{(m-1)!\,(n-1)!} \)

Teacher's Note:
1. m streets running North–South create m − 1 gaps to cross going East; similarly n − 1 going South.
2. A shortest route never goes West or North, so it is a word made of E's and S's.

Answer: (d) 6 and 6

Explanation:
1. With x Algebra and (12 − x) Calculus books, selections = \( {}^{x}C_{3} \) × \( {}^{12-x}C_{3} \).
2. x = 3: 1 × 84 = 84; x = 4: 4 × 56 = 224; x = 5: 10 × 35 = 350; x = 6: 20 × 20 = 400.
3. Greatest when x = 6: 6 and 6.

Teacher's Note:
1. A quick table of values is the safest way to compare such products.
2. The product is symmetric in x and 12 − x, so the balanced split gives the maximum.

Answer: (a) number of possible subsets of r members from a set of n distinct members.; (b) number of possible binary messages of length n with exactly r 1's.; (d) number of ways of selecting r things out of n different things when a particular thing is always included plus the number of ways of selecting r things out of n, when a particular thing is always excluded.

Explanation:
1. (a) True: by definition.
2. (b) True: choose the r positions of the 1's out of n.
3. (c) False: such paths use r + n steps, giving \( {}^{n+r}C_{r} \), not \( {}^{n}C_{r} \).
4. (d) True: \( {}^{n-1}C_{r-1} + {}^{n-1}C_{r} = {}^{n}C_{r} \).

Teacher's Note:
1. Many different counting problems reduce to choosing r positions out of n.
2. Statement (d) is the standard identity behind Pascal's triangle.

Answer: (b) Number of ways in which a multiple choice question containing 10 alternatives with one or more than one correct alternatives, can be answered.; (c) Number of ways in which it is possible to draw a sum of money with 10 coins of different denominations taken some or all at a time.

Explanation:
1. Answer sequences: 2¹⁰; excluding the all-correct one, sequences with incorrect answers = 2¹⁰ − 1.
2. (a) 2¹⁰ − 2; (b) non-empty subsets of 10 alternatives: 2¹⁰ − 1; (c) non-empty subsets of 10 coins: 2¹⁰ − 1.
3. (d) 10 identical things can be selected in 10 ways (1 to 10).
4. So (b) and (c) are correct.

Teacher's Note:
1. Each distinct object (question, alternative, coin) gives a yes/no choice, leading to 2¹⁰.
2. Identical objects only differ by how many are taken, which is why (d) is just 10.

Answer: (b) \( {}^{2n}P_{n} \); (c) (n + 1) (n + 2) (n + 3) … (n + n)

Explanation:
1. Product = \( \prod_{k=1}^{n} 2(2k-1) = 2^n \cdot 1 \cdot 3 \cdot 5 \cdots (2n-1) \).
2. \( 1 \cdot 3 \cdot 5 \cdots (2n-1) = \frac{(2n)!}{2^n\,n!} \), so the product = \( \frac{(2n)!}{n!} = {}^{2n}P_{n} \).
3. \( \frac{(2n)!}{n!} = (n+1)(n+2)\cdots(2n) \), so (b) and (c) are correct.

Teacher's Note:
1. The product of the first n odd numbers is (2n)! divided by the product of the even numbers 2ⁿ n!.
2. Check with n = 2: 2 × 6 = 12 = ⁴P₂ = 3 × 4.

Answer: (a) \( {}^{2n}C_{n} \); (b) \( \frac{2}{1} \cdot \frac{6}{2} \cdot \frac{10}{3} \cdots \frac{4n-6}{n-1} \cdot \frac{4n-2}{n} \); (c) \( \frac{n+1}{1} \cdot \frac{n+2}{2} \cdot \frac{n+3}{3} \cdot \frac{n+4}{4} \cdots \frac{2n-1}{n-1} \cdot \frac{2n}{n} \); (d) \( \frac{2^n\,[1 \cdot 3 \cdots (2n-3)(2n-1)]}{n!} \)

Explanation:
1. With p a's and (2n − p) b's the count is \( \frac{(2n)!}{p!\,(2n-p)!} \), greatest when p = n: \( {}^{2n}C_{n} \).
2. \( {}^{2n}C_{n} = \frac{(n+1)(n+2)\cdots(2n)}{n!} \) gives (c); and \( \frac{(2n)!}{n!\,n!} = \frac{2^n\,[1 \cdot 3 \cdots (2n-1)]}{n!} \) gives (d).
3. (b): the product of \( \frac{4k-2}{k} \) for k = 1 to n is \( \frac{2^n\,[1 \cdot 3 \cdots (2n-1)]}{n!} \), the same as (d).
4. Hence all four are correct.

Teacher's Note:
1. The number of arrangements is maximum when the a's and b's are equal in number.
2. Rewrite \( {}^{2n}C_{n} \) in several product forms to compare with the options.

Answer: (c) \( \frac{(n-1)^2}{4} \) if n is odd; (d) \( \frac{n(n-2)}{4} \) if n is even

Explanation:
1. Three numbers a < b < c are in AP exactly when a + c = 2b, i.e. a and c have the same parity; then b is fixed.
2. n even: n/2 odd and n/2 even numbers: \( 2\,{}^{n/2}C_2 = \frac{n(n-2)}{4} \).
3. n odd: \( \frac{n+1}{2} \) odd and \( \frac{n-1}{2} \) even numbers: \( {}^{(n+1)/2}C_2 + {}^{(n-1)/2}C_2 = \frac{(n-1)^2}{4} \).
4. So (c) and (d) are correct.

Teacher's Note:
1. Choosing the two end terms of the same parity fixes the middle term automatically.
2. Check with n = 4: APs (1,2,3), (2,3,4) give 2 = 4·2/4.

Answer: (a) r! · P(n, n − r); (c) n · P(n − 1, n − 1); (d) P(n, n − 1)

Explanation:
1. P(n, n) = n!.
2. (a) \( r! \cdot \frac{n!}{r!} = n! \), true. (b) \( (n-r)\cdot\frac{n!}{(n-r)!} \) is not n! in general, false.
3. (c) n · (n − 1)! = n!, true. (d) \( \frac{n!}{1!} = n! \), true.

Teacher's Note:
1. Write each P(n, r) as \( \frac{n!}{(n-r)!} \) and simplify.
2. P(n, n − 1) = P(n, n) because once n − 1 objects are placed, the last one is forced.

Answer: (c) 560

Explanation:
1. Take A as (0, 0); then P = (3, 2) and C = (8, 5).
2. A to P: 3 horizontal + 2 vertical steps: \( {}^{5}C_{2} \) = 10 routes.
3. P to C: 5 horizontal + 3 vertical steps: \( {}^{8}C_{3} \) = 56 routes.
4. Routes through P = 10 × 56 = 560

Teacher's Note:
1. A shortest route is a sequence of right and up steps, so count by choosing positions of the up steps.
2. Routes through a point = (routes to it) × (routes from it).

Answer: (b) 350

Explanation:
1. With A = (0, 0): P = (3, 2), Q = (4, 2), C = (8, 5).
2. A to P: \( {}^{5}C_{2} \) = 10 routes; then the step P → Q is fixed.
3. Q to C: 4 horizontal + 3 vertical steps: \( {}^{7}C_{3} \) = 35 routes.
4. Routes along PQ = 10 × 1 × 35 = 350

Teacher's Note:
1. Following a particular street segment means passing P and then Q consecutively.
2. Multiply the counts of the independent parts of the route.

Answer: (b) 240

Explanation:
1. With A = (0, 0): P = (3, 2), R = (6, 3), C = (8, 5).
2. A to P: \( {}^{5}C_{2} \) = 10; P to R (3 horizontal, 1 vertical): \( {}^{4}C_{1} \) = 4; R to C (2 horizontal, 2 vertical): \( {}^{4}C_{2} \) = 6.
3. Routes = 10 × 4 × 6 = 240

Teacher's Note:
1. Break the route at each required junction and multiply.
2. R must be reachable from P by right and up steps only, which it is here.

Answer: (a) (a)→(r), (b)→(s), (c)→(q), (d)→(p)

Explanation:
1. (a) Choose any m of the n numbers; their increasing order is fixed: \( {}^{n}C_{m} \) → (r).
2. (b) Each of the n monkeys chooses one of the m men: mⁿ → (s).
3. (c) The (m − 1) green balls make m gaps; place the n red balls in n different gaps: \( {}^{m}C_{n} \) → (q).
4. (d) Each of the m toys goes to one of the n children: n^m → (p).

Teacher's Note:
1. In 'each X gets one Y' problems, the base is the number of Y's and the power is the number of X's.
2. Fixed increasing order turns an arrangement problem into a selection problem.

Answer: 159

Explanation:
1. All 4-digit numbers formed are greater than 1000 (the smallest is 1012).
2. All four digits different (from 0, 1, 2, 3, 4): \( {}^{5}P_{4} \) − \( {}^{4}P_{3} \) = 120 − 24 = 96.
3. Using both 1's and 2 of {0, 2, 3, 4}: \( {}^{4}C_{2} \) × \( \frac{4!}{2!} \) = 72, minus those starting with 0: 3 × \( \frac{3!}{2!} \) = 9, giving 63.
4. Total = 96 + 63 = 159

Teacher's Note:
1. Split by whether the repeated digit 1 is used once or twice.
2. Subtract arrangements with 0 in the first place in every case that uses 0.

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

Explanation:
1. \( {}^{n}P_{r-1} = \frac{n!}{(n-r+1)!} \)
2. \( (n-r+1)\,{}^{n}P_{r-1} = \frac{(n-r+1)\,n!}{(n-r+1)(n-r)!} = \frac{n!}{(n-r)!} \)
3. \( = {}^{n}P_{r} \)

Teacher's Note:
1. Write (n − r + 1)! as (n − r + 1) × (n − r)! so that the extra factor cancels.
2. This identity says that filling the r-th place after r − 1 places are filled leaves n − r + 1 choices.

Answer: (c) 5

Explanation:
1. \( {}^{5}P_{r} = \frac{5!}{(5-r)!} = \frac{120}{(5-r)!} \)
2. So 120/(5 − r)! = 120 ⇒ (5 − r)! = 1 ⇒ 5 − r = 0 or 1, i.e., r = 5 or r = 4.
3. Among the options, r = 5.

Teacher's Note:
1. Since 0! = 1! = 1, both ⁵P₄ and ⁵P₅ equal 120.
2. Arranging 4 of 5 objects or all 5 objects gives the same count because the last place has only one choice.

Answer: (d) 5

Explanation:
1. \( \frac{{}^{n}P_{5}}{{}^{n}P_{3}} = \frac{n!/(n-5)!}{n!/(n-3)!} = (n-3)(n-4) \)
2. (n − 3)(n − 4) = 2 ⇒ n² − 7n + 10 = 0 ⇒ n = 2 or 5
3. n must be at least 5, so n = 5.

Teacher's Note:
1. The ratio of two permutation symbols with the same n is a product of a few consecutive integers.
2. Always reject the root that makes ⁿP₅ meaningless (n ≥ 5 is needed).

Answer: (a) \( {}^{n}P_{r} \)

Explanation:
1. \( {}^{n-1}P_{r-1} = \frac{(n-1)!}{(n-r)!} \)
2. \( n \cdot {}^{n-1}P_{r-1} = \frac{n!}{(n-r)!} = {}^{n}P_{r} \)

Teacher's Note:
1. Think of filling the first place in n ways and the remaining r − 1 places from n − 1 objects.
2. This gives the useful relation ⁿPᵣ = n · ⁿ⁻¹Pᵣ₋₁.

Answer: (b) 6, 2

Explanation:
1. ᵏP₂ = k(k − 1). So (m + n)(m + n − 1) = 56 = 8 × 7 ⇒ m + n = 8.
2. (m − n)(m − n − 1) = 12 = 4 × 3 ⇒ m − n = 4.
3. Solving, m = 6, n = 2.

Teacher's Note:
1. Recognise 56 and 12 as products of two consecutive integers, 8 × 7 and 4 × 3.
2. Check: ⁸P₂ = 56 and ⁴P₂ = 12.