{"id":36624,"date":"2022-11-30T09:57:56","date_gmt":"2022-11-30T04:27:56","guid":{"rendered":"https:\/\/tsboardsolutions.com\/?p=36624"},"modified":"2022-11-30T15:38:35","modified_gmt":"2022-11-30T10:08:35","slug":"maths-2a-permutations-and-combinations-important-questions-very-short-answer-type","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.com\/maths-2a-permutations-and-combinations-important-questions-very-short-answer-type\/","title":{"rendered":"TS Inter Second Year Maths 2A Permutations and Combinations Important Questions Very Short Answer Type"},"content":{"rendered":"
Students must practice these Maths 2A Important Questions<\/a> TS Inter Second Year Maths 2A Permutations and Combinations Important Questions Very Short Answer Type to help strengthen their preparations for exams.<\/p>\n Question 1. Question 2. Question 3. <\/p>\n \u21d2 2n + 2 = 3n – 12 <\/p>\n Question 4. Question 5. Question 6. <\/p>\n Question 7. Question 8. Question 9. <\/p>\n Question 10. Question 11. Question 12. <\/p>\n Question 13. Question 14. <\/p>\n Question 15. <\/p>\n Question 16. Question 17. Question 18. Question 19. <\/p>\n Question 20. Question 21. Question 22. Question 23. <\/p>\n Question 24. Question 25. Question 26. Question 27. <\/p>\n When repetition is not allowed: <\/p>\n The number of four letter words in which atleast one letter repeated is \\(n^r-{ }^n P_r=6^4-{ }^6 P_4\\) = 1296 – 360 = 936.<\/p>\n <\/p>\n Question 28. <\/p>\n Each ring can be rotated in 9 different ways. Question 29. Question 30. Question 31. <\/p>\n Question 32. Question 33. Question 34. Question 35. <\/p>\n Question 36. Question 37. Question 38. Question 39. <\/p>\n Question 40. Question 41. Question 42. <\/p>\n Question 43. Question 44. Question 45. <\/p>\n Question 46. Question 47. Question 48. <\/p>\n Question 49. Question 50. Students must practice these Maths 2A Important Questions TS Inter Second Year Maths 2A Permutations and Combinations Important Questions Very Short Answer Type to help strengthen their preparations for exams. TS Inter Second Year Maths 2A Permutations and Combinations Important Questions Very Short Answer Type Question 1. If = 1680, find n. [Mar.\u201914, May \u201906] … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":[],"categories":[26],"tags":[],"yoast_head":"\nTS Inter Second Year Maths 2A Permutations and Combinations Important Questions Very Short Answer Type<\/h2>\n
\nIf \\({ }^{\\mathrm{n}} \\mathrm{P}_4\\) = 1680, find n. [Mar.\u201914, May \u201906]
\nSolution:
\nGiven \\({ }^{\\mathrm{n}} \\mathrm{P}_4\\) = 1680
\n\\({ }^{\\mathrm{n}} \\mathrm{P}_4\\) = 8 . 7 . 6 . 5
\n= \\({ }^8 \\mathrm{P}_4\\)
\n\u21d2 n = 8<\/p>\n
\nIf \\({ }^{12} \\mathrm{P}_{\\mathrm{r}}\\) = 1320, find r. [March ’09, AP – May 2015]
\nSolution:
\nGiven \\({ }^{12} \\mathrm{P}_{\\mathrm{r}}\\) = 1320
\n\\({ }^{12} \\mathrm{P}_{\\mathrm{r}}\\) = 12 . 11 . 10
\n= \\({ }^{12} \\mathrm{P}_3\\)
\nr = 3.<\/p>\n
\nIf \\({ }^{(n+1)} P_5:{ }^n P_5\\) = 3 : 2, find n.
\nSolution:
\nGiven \\({ }^{(n+1)} P_5:{ }^n P_5\\) = 3 : 2
\n\\(\\frac{(n+1) P_5}{{ }^n P_5}=\\frac{3}{2}\\)<\/p>\n
\n\u21d2 n = 14.<\/p>\n
\nIf \\({ }^n \\mathbf{P}_7\\) = 42 . \\({ }^n \\mathbf{P}_5\\), find n. [TS – Mar. 2017, ’15; May ’12, ’11, ’09, ’07]
\nSolution:
\nGiven \\({ }^n \\mathbf{P}_7\\) = 42 . \\({ }^n \\mathbf{P}_5\\)
\nn (n – 1) (n – 2) (n – 3) (n – 4) (n – 5) (n – 6) = 42 . n(n – 1) (n – 2) (n – 3) (n – 4)
\n(n – 5) (n – 6) = 42
\n\u21d2 n2<\/sup> – 11n + 30 – 42 = 0
\n\u21d2 n2<\/sup> – 11n – 12 = 0
\n\u21d2 n – 12n + n – 12 = 0
\n\u21d2 n (n – 12) + 1 (n – 12) = 0
\n\u21d2 (n – 12) (n + 1) = 0
\n\u21d2 n = 12; n = – 1
\nSince, n is a positive integer, n = 12.<\/p>\n
\nIf \\({ }^{56} \\mathbf{P}_{(r+6)}:^{54} P_{(r+3)}\\) = 30800: 1, find \u2018r\u2019.
\nSolution:
\nGiven \\({ }^{56} \\mathbf{P}_{(r+6)}:^{54} P_{(r+3)}\\) = 30800 : 1
\n\\(\\frac{\\frac{56 !}{(56-r-6) !}}{\\frac{54 !}{(54-r-3) !}}=\\frac{30800}{1}\\)
\n\u21d2 \\(\\frac{\\frac{56 !}{(50-r) !}}{\\frac{54 !}{(51-r) !}}=\\frac{30800}{1}\\)
\n\u21d2 \\(\\frac{\\frac{56 \\cdot 55 \\cdot 54 !}{(50-r) !}}{\\frac{54 !}{(51-r)(50-r) !}}=\\frac{30800}{1}\\)
\n\u21d2 56 55(51 – r) = 30800
\n\u21d2 51 – r = 10
\n\u21d2 r = 41.<\/p>\n
\nIf \\({ }^{12} \\mathbf{P}_5+5 \\cdot{ }^{12} \\mathbf{P}_4={ }^{13} \\mathbf{P}_{\\mathrm{r}}\\), find \u2018r\u2019. [TS – May 2015]
\nSolution:
\nGiven \\({ }^{12} \\mathbf{P}_5+5 \\cdot{ }^{12} \\mathbf{P}_4={ }^{13} \\mathbf{P}_{\\mathrm{r}}\\)
\n\\({ }^{(13-1)} P_5+5{ }^{(13-1)} P_{(5-1)}={ }^{13} P_r\\)
\nWe know that,
\n\\({ }^n P_r={ }^{(n-1)} P_r+r^{(n-1)} P_{r-1}\\)
\n\u2234 r = 5
\n(or) Given \\({ }^{12} \\mathrm{P}_5+5 \\cdot{ }^{12} \\mathrm{P}_4={ }^{13} \\mathrm{P}_{\\mathrm{r}}\\)
\n12 . 11 . 10 . 9 . 8 + 5 . 12 . 11 . 10 . 9 = \\({ }^13 \\mathrm{P}_{\\mathrm{r}}\\)
\n95040 + 59400 = \\({ }^13 \\mathrm{P}_{\\mathrm{r}}\\)
\n\u21d2 154440 = \\({ }^13 \\mathrm{P}_{\\mathrm{r}}\\)
\n13 . 12 . 11 . 10 . 9 = \\({ }^13 \\mathrm{P}_{\\mathrm{r}}\\)
\n\u21d2 \\({ }^{13} \\mathrm{P}_5\\) = \\({ }^13 \\mathrm{P}_{\\mathrm{r}}\\)
\n\u21d2 r = 5.<\/p>\n
\nIf \\({ }^{\\mathrm{n}} \\mathrm{C}_4\\) = 210, find n. [TS – Mar.2019]
\nSolution:
\nGiven \\({ }^{\\mathrm{n}} \\mathrm{C}_4\\) = 210
\n\u21d2 \\(\\frac{n(n-1)(n-2)(n-3)}{1 \\cdot 2 \\cdot 3 \\cdot 4}\\) = 210
\n\u21d2 n (n – 1) (n -2) (n -3) = 5040
\nn (n – 1) (n – 2) (n – 3) (n – 4) = 10 . 9 . 8 . 7
\n\u2234 n = 10.<\/p>\n
\nIf \\({ }^{12} \\mathrm{C}_{\\mathrm{r}}\\) = 495, find the possible values of \u2018r\u2019.
\nSolution:
\nGiven \\({ }^{12} \\mathrm{C}_{\\mathrm{r}}\\) = 495 = 11 . 9 . 5
\n= \\(\\frac{11 \\cdot 10 \\cdot 9 \\cdot 5}{10}=\\frac{11 \\cdot 10 \\cdot 9}{1 \\cdot 2}=\\frac{12 \\cdot 11 \\cdot 10 \\cdot 9}{1 \\cdot 2 \\cdot 12}\\)
\n= \\(\\frac{12 \\cdot 11 \\cdot 10 \\cdot 9}{1 \\cdot 2 \\cdot 3 \\cdot 4}={ }^{12} \\mathrm{C}_4 \\text { (or) }{ }^{12} \\mathrm{C}_8\\)
\n\u2234 r = 4 (or) 8.<\/p>\n
\nIf 10. \\({ }^n c_2\\) = 3 . \\({ }^{n+1} C_3\\), find \u2019n\u2019. [May ’12], [AP – Mar. 2015]
\nSolution:
\nGiven 10. \\({ }^n c_2\\) = 3 . \\({ }^{n+1} C_3\\)
\n\u21d2 10 . \\(\\frac{\\mathrm{n}(\\mathrm{n}-1)}{1 \\cdot 2}\\) = 3. \\(\\frac{(n+1) n(n-1)}{1 \\cdot 2 \\cdot 3}\\)
\n\u21d2 n + 1 = 10
\n\u21d2 n = 9.<\/p>\n
\nIf \\({ }^{{ }^n} P_{\\mathbf{r}}\\) = 5040 and \\({ }^n C_{\\mathbf{r}}\\) = 210 find n and r. [AP – Mar. \u201817, \u201816; Board Paper]
\nSolution:
\n\\({ }^{{ }^n} P_{\\mathbf{r}}\\) = 5040 ……………(1)
\n\\({ }^n C_{\\mathbf{r}}\\) = 210 ……………..(2)
\n\\(\\frac{(2)}{(1)} \\Rightarrow \\frac{{ }^n C_r}{{ }^n P_r}=\\frac{210}{5040}\\)
\n\\(\\frac{\\frac{n !}{(n-r) ! r !}}{\\frac{n !}{(n-r) !}}=\\frac{1}{24} \\Rightarrow \\frac{1}{r !}=\\frac{1}{4 !}\\)
\nr = 4
\nNow, substituting r = 4 in equation (1)
\n\\({ }^n \\mathrm{p}_4\\) = 5040
\n\\({ }^n \\mathrm{p}_4\\) = 10 . 9 . 8 . 7
\n\\({ }^n \\mathrm{p}_4={ }^{10} \\mathrm{P}_4\\)
\nn = 10.<\/p>\n
\nIf \\({ }^n C_4={ }^n C_6\\), find \u2018n\u2019.
\nSolution:
\nGiven \\({ }^n C_4={ }^n C_6\\)
\nIf \\({ }^n C_r={ }^n C_s\\)
\n\u21d2 n = r + s (or) r = s
\nNow, n = r + s
\n\u21d2 n = 4 + 6
\n\u21d2 n = 10.<\/p>\n
\nIf \\({ }^{15} \\mathrm{C}_{2 \\mathrm{r}-1}={ }^{15} \\mathrm{C}_{2 \\mathrm{r}+4}\\) find \u2019r\u2019. [Mar. ’14, ’05]
\nSolution:
\nGiven \\({ }^{15} \\mathrm{C}_{2 \\mathrm{r}-1}={ }^{15} \\mathrm{C}_{2 \\mathrm{r}+4}\\)
\nIf \\({ }^n C_r={ }^n C_s\\)
\n\u21d2 n = r + s (or) r = s
\n\u21d2 2r – 1 = 2r + 4
\n\u21d2 – 1 \u2260 4
\nIt is not possible.
\nn = r + s
\n\u21d2 15 = 2r – 1 + 2r + 4
\n\u21d2 15 = 4r + 3
\n\u21d2 4r = 12
\n\u21d2 r = 3.<\/p>\n
\nIf \\({ }^{12} \\mathrm{C}_{\\mathrm{r}+1}={ }^{12} \\mathrm{C}_{3 \\mathrm{r}-5}\\), find \u2018r\u2019. [TS- Mar. 2016; March ’08]
\nSolution:
\nGiven \\({ }^{12} \\mathrm{C}_{\\mathrm{r}+1}={ }^{12} \\mathrm{C}_{3 \\mathrm{r}-5}\\)
\nIf \\({ }^n C_r={ }^n C_s\\)
\n\u21d2 n = r + s (or) r = s
\n\u21d2 r + 1 = 3r – 5
\n\u21d2 2r = 6
\n\u21d2 r = 3
\n(or)
\nn = r + s
\n\u21d2 12 = r – 1 + 3r – 5
\n\u21d2 12 = 4r – 4
\n\u21d2 4r = 16
\n\u21d2 r = 4.
\n\u2234 r = 3 (or) 4.<\/p>\n
\nIf \\({ }^9 C_3+{ }^9 C_5={ }^{10} C_{r^{\\prime}}\\) then find \u2018r\u2019.
\nSolution:
\nGiven \\({ }^9 C_3+{ }^9 C_5={ }^{10} C_{r^{\\prime}}\\)<\/p>\n
\nFind the number of ways of forming a committee of 5 members from 6 men and 3 ladies.
\nSolution:
\nNumber of ways of forming a committee of 5 members from 6 men and 3 ladies is
\n\\({ }^9 \\mathrm{C}_5=\\frac{9 \\cdot 8 \\cdot 7 \\cdot 6 \\cdot 5}{1 \\cdot 2 \\cdot 3 \\cdot 4 \\cdot 5}\\) = 126.<\/p>\n
\nIf \\({ }^n C_5={ }^n C_6\\) then find \\({ }^{13} \\mathrm{C}_{\\mathrm{n}}\\). [AP – Mar.2019] [TS-Mar. \u201818; May\u201914, \u201810, March ’13].
\nSolution:
\nGiven \\({ }^n C_5={ }^n C_6\\)
\nIf \\({ }^{13} \\mathrm{C}_{\\mathrm{n}}\\)
\nr = s (or) n = r + s
\nn = r + s = 5 + 6 =11
\nNow,
\n\\({ }^{13} C_n={ }^{13} C_2\\)
\n= \\({ }^{13} \\mathrm{C}_2=\\frac{13 \\cdot 12}{2 \\cdot 1}\\)
\n= 13 . 6 = 78.<\/p>\n
\nProve that \\({ }^{10} \\mathrm{C}_3+{ }^{10} \\mathrm{C}_6={ }^{11} \\mathrm{C}_4\\).
\nSolution:
\nGiven \\({ }^{10} \\mathrm{C}_3+{ }^{10} \\mathrm{C}_6={ }^{11} \\mathrm{C}_4\\)
\nL.H.S: \\({ }^{10} \\mathrm{C}_3+{ }^{10} \\mathrm{C}_6={ }^{10} \\mathrm{C}_3+{ }^{10} \\mathrm{C}_4\\)
\n[\u2235 \\({ }^n C_r={ }^n C_{n-r}\\)]
\n= \\(={ }^{10} \\mathrm{C}_4+{ }^{10} \\mathrm{C}_{4-1}={ }^{(10+1)} \\mathrm{C}_4\\)
\n[\u2235 \\({ }^n C_r+{ }^n C_{n-r}={ }^{(n+1)} C_r\\)]
\n= \\({ }^{11} \\mathrm{C}_4\\)
\n= R.H.S<\/p>\n
\nIf \\({ }^{12} C_{s+1}={ }^{12} C_{2 s-5}\\) find \u2018s\u2019. [Mar. ’11]
\nSolution:
\nGiven \\({ }^{12} C_{s+1}={ }^{12} C_{2 s-5}\\)
\nIf \\({ }^n C_r={ }^n C_s\\)
\n\u21d2 n = r + s (or) r = s
\n\u21d2 r = s
\n\u21d2 s + 1 = 2s – 5
\n\u21d2 s = 6 (or)
\nn = r + s
\n12 = s + 1 + 2s – 5
\n12 = 3s – 4 = 3s = 16
\ns = \\(\\frac{16}{3}\\)
\nSince s is integer, s = 6.<\/p>\n
\nIf \\({ }^n C_{21}={ }^n C_{27}\\) find \\({ }^{50} \\mathrm{C}_{\\mathrm{n}}\\).
\nSolution:
\nGiven \\({ }^n C_{21}={ }^n C_{27}\\)
\nIf \\({ }^n C_r={ }^n C_s\\)
\n\u21d2 n = r + s (or) r = s
\n\u21d2 n = r + s
\n\u21d2 n = 21 + 27
\n\u21d2 n = 48
\nNow, \\({ }^{50} \\mathrm{C}_{\\mathrm{n}}={ }^{50} \\mathrm{C}_{48}\\)
\n= \\({ }^{50} \\mathrm{C}_2=\\frac{50 \\cdot 49}{2 \\cdot 1}\\)
\n= 25 . 49 = 1225.<\/p>\n
\nFind the number of positive divisors of 1080. [AP – May, Mar. 2016; May \u201813]
\nSolution:
\n1080 = 23<\/sup> \u00d7 33<\/sup> \u00d7 51<\/sup>
\nThe number of positive divisors of 1080 = (3 + 1) (3 + 1) (1 + 1)
\n= 4 . 4 . 2 = 32.<\/p>\n
\nFind the value of \\({ }^{10} \\mathrm{C}_5+2 \\cdot{ }^{10} \\mathrm{C}_4+{ }^{10} \\mathrm{C}_3\\). [AP – Mar. \u201818; TS- Mar. 2017; March \u201810]
\nSolution:
\nGiven \\({ }^{10} \\mathrm{C}_5+2 \\cdot{ }^{10} \\mathrm{C}_4+{ }^{10} \\mathrm{C}_3\\)
\n= \\(\\left({ }^{10} \\mathrm{C}_5+{ }^{10} \\mathrm{C}_4\\right)+\\left({ }^{10} \\mathrm{C}_4+{ }^{10} \\mathrm{C}_3\\right)\\)
\n= \\({ }^{11} C_5+{ }^{11} C_4\\)
\n= \\({ }^{12} \\mathrm{C}_5=\\frac{12 \\cdot 11 \\cdot 10 \\cdot 9 \\cdot 8}{1 \\cdot 2 \\cdot 3 \\cdot 4 \\cdot 5}\\)
\n= 792<\/p>\n
\nFind the number of injections ofaset A with 5 elements to a set B with 7 elements.
\nSolution:
\nIf a set A has m elements and the set B has n elements, then the number of injections from A into B is \\(\\mathrm{n}_{\\mathrm{m}}\\) if m \u2264 n and 0 if m > n.
\nGiven n = 7, m = 5
\n\u2234 The number of injections from A to B is \\({ }^7 \\mathrm{P}_5\\) = 7. 6 . 5 . 4 . 3 = 2520.<\/p>\n
\nFind the number of ways in which 4 letters can be put in 4 addressed envelopes so that no letter goes into the envelope meant for it. [TS -May 2016]
\nSolution:
\nThe number of derangements of n distinct things is
\n\\(\\mathrm{n} !\\left(\\frac{1}{2 !}-\\frac{1}{3 !}+\\frac{1}{4 !}-\\frac{1}{5 !}+\\ldots \\ldots+(-1)^{\\mathrm{n}} \\frac{1}{\\mathrm{n} !}\\right)\\)
\nRequired number of ways is \\(4 !\\left(\\frac{1}{2 !}-\\frac{1}{3 !}+\\frac{1}{4 !}\\right)\\) = 12 – 4 + 1 = 9.<\/p>\n
\nA man has 4 sons and there are 5 schools within his reach. In how many ways can he admit his sons in the schools so that no two of them will be in the saine school?
\nSolution:
\nThe number of ways of admitting four sons into five schools if no two of them will be in the same school.
\n\\({ }^n P_r={ }^5 P_4\\) = 5 . 4 . 3 . 2 = 120.<\/p>\n
\nIf there are 25 railway stations on a railway line, how many types of single second class tickets must be printed, so as to enable a passenger to travel from one station to another?
\nSolution:
\nNumber of stations on a railway line = 25
\nNum ber of single different second class tickets must be printed so as to enable a passenger to travel from one station to another = Number of ways of arranging two station names out of 25 station names = \\({ }^{25} \\mathrm{P}_2\\)
\n= 25 . 24 = 600.<\/p>\n
\nIn a class, there are 30 students on the new year day, every student posts a greeting card to all his\/her classmates. Find the total number of greeting cards posted by them.
\nSolution:
\nTotal number of students in a class = 30.
\nTotal number of greeting cards posted by all the students to their classmates = number of ways of arranging names of two students from 30 names of students
\n= \\({ }^n P_r={ }^{30} P_2\\)
\n= 30 \u00d7 29 = 870.<\/p>\n
\nFind the number of 4 letter words that can be formed using the letters of the word PISTON in which atleast one letter is repeated. [AP – Mar. 2015]
\nSolution:
\nThe given word has six letters.
\nWhen repetition is allowed:
\nThe number of four letter words that can be formed using these six letters when repetition is allowed is \\(n^r=6^4\\) = 6 . 6 . 6 . 6 = 1296.<\/p>\n
\nThe number of four letter words that can be formed using these letters of the six letters when repitition is not allowed is
\n\\({ }^n \\mathrm{P}_{\\mathrm{r}}={ }^6 \\mathrm{P}_4\\) = 360<\/p>\n
\nA number lock has 3 rings and each ring has 9 digits 1, 2, 3, ……………, 9. Find the maximum number of unsuccessful attempts that can be made by a persoti who tries to open the lock without knowing the key code.
\nSolution:<\/p>\n
\nThe total no. of different ways in which three kings can be rotated is = 93<\/sup> = 9 . 9 . 9 = 729
\nOut of these attempts, only one attempt is successful attempt.
\nTherefore, the maximum no. of unsuccessful attempts is 729 – 1 = 728.<\/p>\n
\nFind the number of functions from a set A containing 5 elements into a set B containing 4 elements.
\nSolution:
\nGiven number of A = n(A) = m = 5
\nnumber of B = n(B) = n = 4
\nSet A contains 5 elements and set B contains 4 elements.
\nThe total number of functions from set A containing m elements to set B containing n elements is \\(n^m=4^5\\) = 1024.<\/p>\n
\nFind the number of bijections from a set A contaIning 7 elements onto itself.
\nSolution:
\nGiven n(A) = n = 7.
\n\u2234 The number of bijections from set A with n elements to set B with same number of elements \u2018n\u2019, A is n!.
\nThe number of bijections from set A with 7 elements onto itself = 7!
\n= 7. 6 . 5 . 4 . 3 . 2 . 1 = 5040.<\/p>\n
\nFind the number of ways of arranging 7 persons around a circle.
\nSolution:
\nGiven number of persons, n = 7.
\n\u2234 The number of ways of arranging 7 persons around a circle is
\n(n – 1)! = (7 – 1)! = 6!
\n= 6 . 5 . 4 . 3 . 2 . 1 = 720.<\/p>\n
\nFind the number of ways of preparing a chain with 6 different coloured beads. [TS – Mar. \u201819, 16; March \u201808]
\nSolution:
\nNeglecting the directions of beads in the chain.
\nNumber of ways of preparing a chain with 6 different coloured beads = \\(\\frac{(n-1) !}{2}\\)
\n= \\(\\frac{(6-1) !}{2}=\\frac{5 !}{2}=\\frac{5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1}{2}\\) = 60.<\/p>\n
\nFind the number of ways of arranging the Chief Minister and 10 Cabinet Ministers at a circular table so that the Chief Minister always sits in a particular seat.
\nSolution:
\nTotal number of persons = 11
\nChief Minister can sit in a particular seat in one way.
\nNow, remaining positions are well defined relative to Chief Minister.
\nHence, the remaining can sit in 10 places in 10! ways.
\n\u2234 The number of required arrangements = 10! \u00d7 1
\n= 10 . 9 . 8 . 7 . 6 . 5 . 4 . 3 . 2 . 1 = 3628800.<\/p>\n
\nFind the number of ways of arranging the letters of the word a4<\/sup> b3<\/sup> c5<\/sup> in its expanded form.
\nSolution:
\nThe expanded form of a4<\/sup> b3<\/sup> c5<\/sup> is a . a . a . a . b . b . b . c . c . c . c . c.
\nThis word has 12 letters in which there are 4a’s, 3b’s, 5c’s.
\n\u2234 They can be arranged in \\(\\frac{12 !}{4 ! 3 ! 5 !}\\) ways.<\/p>\n
\nThere are 4 copies (alike) each of 3 different books. Find the number of ways of arranging these 12 books in a shelf in a single row.
\nSolution:
\nWe have 12 books in which 4 books are alike of 1 kind. 4 books are alike of 2nd kind and 4 books are alike of 3rd kind.
\nHence, they can be arranged in a shelf in a row in \\(\\frac{12 !}{4 ! 4 ! 4 !}\\) ways = \\(\\frac{12 !}{(4 !)^3}\\) ways.<\/p>\n
\nFind the number of ways of arranging the letters of the word INDEPENDENCE. [TS – May 2016; March \u201909, May \u201913].
\nSolution:
\nGiven word is INDEPENDENCE
\nThere are 12 letters in INDEPENDENCE, in which there are 3 Ns are alike. 2 D\u2019s are alike, 4 E\u2019s are alike, and rest are different.
\n\u2234 The no. of required arrangements = \\(\\frac{12 !}{3 ! \\cdot 2 ! \\cdot 4 !}\\)<\/p>\n
\nFind the number of ways of arranging the letters of the word MATHEMATICS. [AP & TS – Mar. ’18; May ’97, ’10, March ’11, ’06]
\nSolution:
\nGiven word is MATHEMATICS.
\nThe word MATHEMATICS contains 11 letters in which there are , 2 M’s are alike, 2 A’s are alike, 2 T’s are alii ice and rest are different.
\n\u2234 The no. of required arran igements 11! = \\(\\frac{11 !}{2 ! \\cdot 2 ! \\cdot 2 !}\\).<\/p>\n
\nFind the number of ways of arranging the letters of the word INTERMEDIATE. [AP -Mar. ’19; May 2016; May ’14, Board Paper]
\nSolution:
\nGiven word is INTERMEDIATE.
\nThe word INTERMEDIATE contains 121 letters in which there are 2 I’s are alike, 2 T\u2019s are alike, 3 E’s are alike and rest are different.
\n\u2234 The no. of required arrangements = \\(\\frac{12 !}{2 ! \\cdot 2 ! \\cdot 3 !}\\).<\/p>\n
\nFind the number of 7 digit numbers that can be fonned using 2, 2, 2, 3, 3, 4 4.
\nSolution:
\nIn the given seven digits, there are three 2\u2019s, two 3\u2019s, two 4\u2019s.
\n\u2234 The number of seven digited numbers that can be formed using th given digits = \\(\\frac{7 !}{3 ! \\cdot 2 ! \\cdot 2 !}\\).<\/p>\n
\nFind the number of ways of selecting 7 members from a continent of 10 soldiers.
\nSolution:
\nThe number of ways of selecting 7 members out of 10 soldiers is \\({ }^{10} \\mathrm{C}_7\\) = 120.<\/p>\n
\nA set A has 8 elements, find the number of subsets of A, containing atleaist 6 elements.
\nSolution:
\nThe no. of subsets of A containing atlast 6 elements then the number of subsets of A containing 6 or 7 or 8 elements.
\nNumber of subsets of A containing exactly 6 elements = \\({ }^8 \\mathrm{C}_6\\)
\nNumber cf subsets of A containing exactly 7 elements = \\({ }^8 \\mathrm{C}_7\\)
\nNumber of :ubsets of A containing exactly 8 elements = \\({ }^8 \\mathrm{C}_8\\)
\nThe nurrber of subsets of A containing at least 6 elements = \\({ }^8 \\mathrm{C}_6+{ }^8 \\mathrm{C}_7+{ }^8 \\mathrm{C}_8\\)
\n= \\({ }^8 \\mathrm{C}_2+{ }^8 \\mathrm{C}_1+{ }^8 \\mathrm{C}_0\\)
\n= \\(\\frac{8 \\cdot 7}{2 \\cdot 1}\\) + 8 + 1
\n= 28 + 8 – 1 = 37.<\/p>\n
\nFind the number of ways of selecting 4 boys and 3 girls from a group of 8 boys and 5 girls. [TS – Mar. 2015]
\nSolution:
\n4 boys can be selected from the given boys in \\({ }^8 \\mathrm{C}_4\\) ways.
\n3 girls can be seleted from the given 5 girls in \\({ }^5 \\mathrm{C}_3\\) ways.
\n\u2234 The required number of selections is \\({ }^8 \\mathrm{C}_4 \\cdot{ }^{! 5} \\mathrm{C}_3\\) = \\(\\frac{8 \\cdot 7 \\cdot 6 \\cdot 5}{4 \\cdot 3 \\cdot 2 \\cdot 1} \\frac{5 \\cdot 4}{2 \\cdot 1}\\) = 700.<\/p>\n
\nIf there are 5 alIke pens, 6 alike pencils and 7 alike erasers, find the number of ways of selecting any number of (one or
\nmore ) things out of them.
\nSolution:
\nThe required number of ways is (p + 1) (q + 1) (r + 1) – 1
\n= (5 + 1) (6 + 1) (7 + 1) – 1
\n= 6 . 7 . 8 – 1
\n= 336 – 1 = 335.<\/p>\n
\nTo pass an examination a student has to pass in each of the three papers. In how many ways can a student fail in the exaimination? [TS- May 2015]
\nSolution:
\nFor each of the three papers there are two choices P or F.
\nThere are 23<\/sup> = 8 choices.
\nBut a student passes only il he\/she passes in all papers.
\n\u2234 Required number of ways = 23<\/sup> – 1 = 7.<\/p>\n
\nIn a class, there are 30 students. If each student plays a chess game with each of the other student, then find the total number of chess games played by them.
\nSolution:
\nNumber of students in a class is 30.
\nSince each student plays a chess game with each of the student the total number of games played by them = \\({ }^{30} \\mathrm{C}_2\\)
\n= \\(\\frac{30 \\cdot 29}{2}\\) = 435.<\/p>\n
\nFind the number of diagonals of a polygon with 12 sides. [AP – May 2015]
\nSolution:
\nNumber of sides of a polygon = 12
\nNumber of diagonals of a n – sided polygon = \\({ }^n C_2\\) – n
\n\u2234 Number of diagonals of 12 sided polygon = \\({ }^{12} \\mathrm{C}_2\\) – 12 = 54.<\/p>\n
\nIf \\({ }^{\\mathrm{n}} \\mathrm{P}_3\\) = 1320, find n. [May \u201808, March \u201805]
\nSolution:
\n12<\/p>\n
\nIf \\({ }^{(n+1)} P_5:{ }^n P_6\\) = 2 : 7, find n. [March \u201810, \u201807]
\nSolution:
\n11<\/p>\n
\nIf \\({ }^{18} P_{(r-1)}:^{17} P_{(r-1)}\\) = 9 : 7, find \u2018r\u2019.
\nSolution:
\n5<\/p>\n
\nFInd the number of different chains that can be prepared using 7 different coloured beads. [AP – Mar. 2017]
\nSolution:
\n360<\/p>\n","protected":false},"excerpt":{"rendered":"