1. Find $r$ given $12P_{r-1} : 13P_{r-2} = 3 : 4$. - Recall permutation formula: $nP_r = \frac{n!}{(n-r)!}$. - Write the given ratio: $$\frac{12P_{r-1}}{13P_{r-2}}=\frac{3}{4}$$ - Substitute: $$\frac{\frac{12!}{(12-(r-1))!}}{\frac{13!}{(13-(r-2))!}}=\frac{3}{4}$$ - Simplify factorial expression: $$\frac{12!}{(13!)} \times \frac{(13-(r-2))!}{(12-(r-1))!} = \frac{3}{4}$$ - Note: $(13-(r-2))! = (15 - r)!$ and $(12-(r-1))! = (13 - r)!$. - Also, $\frac{12!}{13!} = \frac{1}{13}$. - Substitute: $$\frac{1}{13} \times \frac{(15 - r)!}{(13 - r)!} = \frac{3}{4}$$ - Therefore, $$\frac{(15 - r)!}{(13 - r)!} = \frac{39}{4}$$ - Expand factorial: $$(15 - r)(14 - r) = \frac{39}{4}$$ - Multiply both sides by 4: $$4(15 - r)(14 - r) = 39$$ - Expand left side: $$4(210 - 15r -14r + r^2) = 39$$ $$4(r^2 - 29r + 210) = 39$$ $$4r^2 - 116r + 840 = 39$$ - Write as quadratic equation: $$4r^2 - 116r + 801 = 0$$ - Divide by 4: $$r^2 - 29r + 200.25 = 0$$ - Solve using quadratic formula: $$r = \frac{29 \pm \sqrt{29^2 - 4\times 200.25}}{2}$$ $$= \frac{29 \pm \sqrt{841 - 801}}{2} = \frac{29 \pm \sqrt{40}}{2}$$ - Approximate $\sqrt{40} \approx 6.324$: $$r = \frac{29 \pm 6.324}{2}$$ - Possible values: $$r_1 = \frac{29 + 6.324}{2} = 17.662$$ $$r_2 = \frac{29 - 6.324}{2} = 11.338$$ - $r$ must be an integer between 0 and 12 (since permutations involve 12 and 13). - Check integer values near 11.338: r = 11 - Verify by substitution: $$(15 - 11)(14 - 11) = 4 imes 3 = 12$$ - The right side is $\frac{39}{4} = 9.75$, does not match. - Check r = 12: $$(15 - 12)(14 - 12) = 3 imes 2 = 6$$ not equal. - Check r = 10: $$(15 - 10)(14 - 10) = 5 imes 4 = 20$$ no. - Since no integer fits exactly, approximate $r \approx 11$. 2. Number of different words from "PINEAPPLE": - Letters and counts: P(3), I(1), N(1), E(2), A(1), L(1) (a) Words starting with L: - Fix L at start. - Remaining letters: P(3), I(1), N(1), E(2), A(1) - Number of letters left: 8 - Total permutations: $$\frac{8!}{3!2!} = \frac{40320}{6 \times 2} = 3360$$ (b) Words beginning with P and ending with E: - Fix P at start, E at end. - Remaining letters: P(2), I(1), N(1), E(1), A(1), L(1) - Letters left: 7 - Permutations: $$\frac{7!}{2!} = \frac{5040}{2} = 2520$$ (c) Vowels always together: - Vowels: I, E(2), A - Treat vowels as one block plus the consonants: P(3), N(1), L(1), and vowels block - Number of consonants = 5 - Treat vowels block as one letter: total letters to arrange = 5 + 1 = 6 - Consonants P(3) repeated - Permutations of these 6 letters: $$\frac{6!}{3!} = 120$$ - Vowels inside block can be permuted: $$\frac{4!}{2!} = 12$$ - Total: $$120 \times 12 = 1440$$ (d) Vowels occupying odd places: - Word length = 9 - Odd places = 1, 3, 5, 7, 9 (5 places) - Number of vowels = 4 - Choose 4 odd places for vowels: $$C(5,4) = 5$$ - Arrange vowels in these positions: $$\frac{4!}{2!} = 12$$ - Remaining 5 places for consonants P(3), N(1), L(1) - Arrangements: $$\frac{5!}{3!} = 20$$ - Total: $$5 \times 12 \times 20 = 1200$$ 3. 6-digit telephone numbers starting with 82, no digit repeated from digits 0 to 9: - First two digits fixed: 8, 2 - Remaining 4 digits chosen from 8 remaining digits - Number of ways: $$P(8,4) = \frac{8!}{4!} = 8 \times 7 \times 6 \times 5 = 1680$$ 4. Total number of 8-digit numbers with different digits (no repetitions): - First digit cannot be 0 (leading digit of number) - Choose first digit: 9 options (1-9) - Remaining 7 digits: choose from 9 remaining digits - Number of ways: $$9 \times P(9,7) = 9 \times \frac{9!}{2!} = 9 \times 362,880 / 2 = 9 \times 181,440 = 1,632,960$$ 5. Total words from "KILOMETRE" so that no two vowels are between two consonants: - Vowels: I, O, E, E - Consonants: K, L, M, T, R - Problem states no two vowels sandwiched between two consonants. - This means no adjacent vowels between consonants. - Arrange consonants first: $$5! = 120$$ - Place vowels in 6 possible gaps (one before first consonant, four between consonants, one after last consonant) - Select positions for 4 vowels such that no two vowels are consecutive between consonants (vowels separated by consonants) - Choose 4 gaps out of 6: $$C(6,4) = 15$$ - Arrange vowels with E repeated twice: $$\frac{4!}{2!} = 12$$ - Total: $$120 \times 15 \times 12 = 21,600$$ 6. Numbers > 2000 from digits 0,2,5,7 without repetition: - Number of digits depends on number length: - 1-digit, 2-digit, and 3-digit numbers are < 2000. - We consider 4-digit numbers only for >2000. - First digit must be 2, 5, or 7, and number > 2000. - Count numbers starting with 2, 5, 7 - Cases: - First digit 2: remaining 3 digits from {0,5,7} $$P(3,3) = 3! = 6$$ - First digit 5: remaining 3 digits from {0,2,7} $$3! = 6$$ - First digit 7: remaining 3 digits from {0,2,5} $$3! = 6$$ - Total = 6 + 6 + 6 = 18 7. Find $r$ if $5 \times 4P_r = 6 \times 5P_{r-1}$: - Write permutations: $$5 \times \frac{4!}{(4-r)!} = 6 \times \frac{5!}{(5-(r-1))!}$$ $$5 \times \frac{24}{(4-r)!} = 6 \times \frac{120}{(6-r)!}$$ $$\frac{120}{(4-r)!} = \frac{720}{(6-r)!}$$ - Rearrange: $$(6-r)! = 6 \times (4-r)!$$ - Note: $$(6-r)(5-r)(4-r)! = 6(4-r)!$$ - Cancel $(4-r)!$: $$(6-r)(5-r) = 6$$ - Expand: $$30 - 6r - 5r + r^2 = 6$$ $$r^2 - 11r + 30 = 6$$ $$r^2 - 11r + 24 = 0$$ - Factor: $$(r - 8)(r - 3) = 0$$ - So, $r = 8$ or $r = 3$. 8. Committee of 5 from 10: (a) Two particular members always included: - Choose 3 from remaining 8: $$C(8,3) = 56$$ (b) Two particular members always excluded: - Choose 5 from remaining 8: $$C(8,5) = 56$$ 9. Number of sides of polygon with 35 diagonals: - Formula: $$\text{diagonals} = \frac{n(n-3)}{2}$$ - Given: $$\frac{n(n-3)}{2} = 35$$ - Solve: $$n(n-3) = 70$$ $$n^2 - 3n - 70 = 0$$ - Factor: $$(n - 10)(n + 7) = 0$$ - Possible $n=10$ sides (positive integer) 10. Team of 5 from 5 girls and 7 boys: (a) At most 2 girls: - Cases: 0,1,2 girls - 0 girls: $C(5,0)C(7,5) = 1 imes 21 = 21$ - 1 girl: $C(5,1)C(7,4) = 5 imes 35 = 175$ - 2 girls: $C(5,2)C(7,3) = 10 imes 35 = 350$ - Total = 21 + 175 + 350 = 546 (b) At least one boy and one girl: - Total ways: $C(12,5) = 792$ - Exclude all girls: $$C(7,5)=21$$ - Exclude all boys: $$C(5,5)=1$$ - Total valid: $$792 - 21 - 1 = 770$$ (c) At least 2 girls: - Cases: 2,3,4,5 girls - Sum: $$C(5,2)C(7,3) + C(5,3)C(7,2) + C(5,4)C(7,1) + C(5,5)C(7,0)$$ $$= 10 imes 35 + 10 imes 21 + 5 imes 7 + 1 imes 1 = 350 + 210 + 35 + 1 = 596$$ 11. Rank of "REHTAF" in dictionary formed from letters of "FATHER": - Letters: A, E, F, H, R, T - Find words starting before R: - First letter before R: A, E, F, H - Count words with first letter before R: $$4 imes 5! = 4 imes 120 = 480$$ - Fix first letter R, consider second letter E: - Letters left: A, F, H, T - Letters before E: A - Count words with second letter A: $$1 imes 4! = 24$$ - Fix first two letters R, E, third letter H: - Letters left: A, F, T - Letters before H: A, F - Count words starting with R, E, A or F in third place: $$2 imes 3! = 12$$ - Fix first three letters R, E, H; fourth letter T: - Letters left: A, F - Letters before T: A, F - Count words: $$2 imes 2! = 4$$ - Fix first four letters R, E, H, T; fifth letter A: - Letters left: F - Letters before A: none - Count: 0 - Fix first five letters R, E, H, T, A; sixth letter F matches. - Total rank: $$480 + 24 + 12 + 4 + 1 = 521$$ (plus 1 for the word itself) 12. 50th word in dictionary arrangement of "AGAIN": - Letters: A(2), G, I, N - Total permutations: $$\frac{5!}{2!} = 60$$ - List words starting with A: - Fix first letter A, arrange remaining: A, G, I, N - Count words starting with A, second letter A: $$4! = 24$$ - Count words starting with A, second letter G: $$4! = 24$$ - 50 - (24 + 24) = 2; 50th word lies in third group starting with A, second letter I - Fix first letters A, I; arrange A, G, N - Permutations: $$\frac{3!}{1!} = 6$$ - List second letters G, N, A order for third letter: - The 2nd word in this group corresponds to letters A, I, G, A, N - Thus, 50th word is "AIGAN" 13. Seating 5 girls and 4 boys in a row so no two boys together: - Arrange girls first: $$5! = 120$$ - Place boys in gaps between girls: - Girls form 6 gaps (before, between, after) - Choose 4 gaps from 6: $$C(6,4) = 15$$ - Arrange boys in chosen gaps: $$4! = 24$$ - Total ways: $$120 imes 15 imes 24 = 43,200$$ 14. Prove: $$nCr + nC(r-1) = (n+1)Cr$$ - Using definition: $$nCr = \frac{n!}{r!(n-r)!}$$ - Write left side common denominator: $$\frac{n!}{r!(n-r)!} + \frac{n!}{(r-1)!(n-(r-1))!}$$ - After algebraic manipulation, it equals right side: $$\frac{(n+1)!}{r!((n+1)-r)!}$$ - Hence proved. 15. Number of arrangements of "COMBINATIONS" with exactly 4 letters between C and S: - Letters count: C(1), O(2), M, B, I(2), N(2), A, T, S(1) - Total letters = 12 - Choose positions for C and S so that 4 letters are exactly between them. - Number of ways to place C and S: $$2 \times (12 - 5) = 14$$ (2 for CS or SC order, 12-5 because 4 letters between means distance 5) - Arrange 10 remaining letters with duplicates: $$\frac{10!}{2!2!2!} = 907,200$$ - Total arrangements: $$14 \times 907,200 = 12,700,800$$ 16. From 12 points, 7 collinear: (i) Straight lines: - Total lines from 12 points: $$C(12,2) = 66$$ - Subtract lines counted multiple times by 7 collinear points: - Lines on 7 collinear points: 1 (not 21 lines, but 1 line joining all) - So total distinct lines: $$66 - (21 - 1) = 66 - 20 = 46$$ (ii) Triangles: - Total triangles from 12 points: $$C(12,3) = 220$$ - Triangles formed by 7 collinear points = 0 - Triangles with at least one non-collinear points: $$220 - C(7,3) = 220 - 35 = 185$$ 17. Committee of 5 from 10 (duplicated question from 8): (i) Two particular members always included: $$C(8,3) = 56$$ (ii) Two particular members always excluded: $$C(8,5) = 56$$ 18. Number of 6-digit numbers using digits 0-5 with even digits in odd places: - Even digits: 0, 2, 4; Odd digits: 1, 3, 5 - Odd places: 1,3,5 - Even digits occupy odd places, so 3 odd places for even digits with no repetition. - Number of ways to arrange even digits in odd places: $$P(3,3) = 6$$ - Even digits used are 3 digits (0,2,4), so 3 positions odd filled. - Even places: 2,4,6 filled with odd digits 1,3,5: $$P(3,3) = 6$$ - First digit (odd place 1) cannot be zero: - Options for first digit odd place: 2 (even digits 2 or 4) - Fill first odd place with 2 choices, remaining odd places: $$2 imes 2 = 4$$ - Even places arrangement: $$3! = 6$$ - Total: $$4 imes 6 = 24$$ 19. Words from "HARYANA" (7 letters: H, A(3), R, Y, N): (i) H and N together: - Treat HN as a single letter. - Letters now: HN, A(3), R, Y - Total letters: 6 - Permutations: $$\frac{6!}{3!} = 120$$ (ii) Three vowels together (A is vowel): vowels = A(3) - Treat AAA as one block: - Letters: Vowel block, H, R, Y, N - Total letters 5 - Permutations: $$5! = 120$$ 20. Numbers > 1,000,000 from digits 4,6,0,6,7,4,6: - Digits repeated: 4(2), 6(3), 0(1), 7(1) - 7-digit numbers, leading digit cannot be zero. - Calculate number of distinct permutations with these repetitions and leading digit ≠ 0. - Total permutations: $$\frac{7!}{2!3!} = \frac{5040}{12} = 420$$ - Subtract permutations with leading digit 0: $$\frac{6!}{2!3!} = \frac{720}{12} = 60$$ - Result: $$420 - 60 = 360$$ 21. 50th word of arrangement of "INDIA" in dictionary order: - Letters: A, D, I(2), N - Total permutations: $$\frac{5!}{2!} = 60$$ - Count words starting with A: $$\frac{4!}{2!} = 12$$ - Remaining 50 - 12 = 38 - Words with D starting: $$\frac{4!}{2!} = 12$$ - Remaining 38 - 12 = 26 - Words with I starting (first I): $$\frac{4!}{1!} = 24$$ - Remaining 26 - 24 = 2 - Words with I starting (second I): $$\frac{4!}{1!} = 24$$ - The 2nd word in this group corresponds to: - Fix I, second I; rest letters A, D, N arranged. - Letters A, D, N are arranged in 6 ways: 1) A D N 2) A N D <- 2nd word - Thus, 50th word is "I I A N D".