{"id":9422,"date":"2023-11-01T16:12:10","date_gmt":"2023-11-01T10:42:10","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=9422"},"modified":"2023-11-03T16:54:12","modified_gmt":"2023-11-03T11:24:12","slug":"ts-inter-2nd-year-maths-2a-solutions-chapter-5-ex-5a","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-inter-2nd-year-maths-2a-solutions-chapter-5-ex-5a\/","title":{"rendered":"TS Inter 2nd Year Maths 2A Solutions Chapter 5 Permutations and Combinations Ex 5(a)"},"content":{"rendered":"

Students must practice this TS Intermediate Maths 2A Solutions<\/a> Chapter 5 Permutations and Combinations Ex 5(a) to find a better approach to solving the problems.<\/p>\n

TS Inter 2nd Year Maths 2A Solutions Chapter 5 Permutations and Combinations Ex 5(a)<\/h2>\n

I.
\nQuestion 1.
\nIf \\({ }^{\\mathrm{n}} \\mathrm{P}_3\\) = 1320, find n.
\nSolution:
\nGiven \\({ }^{\\mathrm{n}} \\mathrm{P}_3\\) = 1320
\n\u21d2 n(n – 1) (n – 2) = 12 \u00d7 11 \u00d7 10
\nOn comparing, we have n = 12.<\/p>\n

Question 2.
\nIf \\({ }^n P_7\\) = 42 P5, find n.
\nSolution:
\nGiven \\({ }^n P_7\\) = 42 . \\({ }^n P_5\\)
\n\u21d2 \\(\\frac{n !}{(n-7) !}=42 \\cdot \\frac{n !}{(n-5) !}\\)
\n\u21d2 \\(\\frac{n !}{(n-7) !}=42 \\cdot \\frac{n !}{(n-5)(n-6)(n-7) !}\\)
\n\u21d2 (n – 5) (n – 6) – 42
\n\u21d2 (n – 5) (n – 6) = 7 \u00d7 6
\nOn comparing largest integers, we have n – 5 = 7
\n\u21d2 n = 12.<\/p>\n

Question 3.
\nIf \\({ }^{(n+1)} P_5:{ }^n P_6\\) = 2 : 7, find n.
\nSolution:
\nGiven \\({ }^{(n+1)} P_5:{ }^n P_6\\) = 2 : 7
\n\u21d2 \\(\\frac{(n+1) !}{(n-4) !}: \\frac{n !}{(n-6) !}\\) = 2 : 7
\n\u21d2 \\(7 \\cdot \\frac{(n+1) !}{(n-4) !}=2 \\cdot \\frac{n !}{(n-6) !}\\)
\n\u21d2 \\(\\frac{7 \\cdot(n+1) \\cdot n !}{(n-4)(n-5)(n-6) !}=2 \\cdot \\frac{n !}{(n-6) !}\\)
\n\u21d2 7(n + 1) = 2 (n – 4) (n – 5)
\n\u21d2 2n2<\/sup> – 25n + 33 = 0
\n\u21d2 (2n – 3) (n – 11) = 0
\n(\u2235 2n – 3 \u2260 0 as an integer)
\n\u21d2 n – 11 = 0
\n\u21d2 n = 11.<\/p>\n

\"TS<\/p>\n

Question 4.
\nIf \\({ }^{12} P_5+5 \\cdot{ }^{12} P_4={ }^{13} P_t\\) find r.
\nSolution:
\nGiven, \\({ }^{12} P_5+5 \\cdot{ }^{12} P_4={ }^{13} P_t\\)<\/p>\n

\"TS<\/p>\n

Alternate method:
\nWe know that
\n\\({ }^n P_r={ }^{(n-1)} P_r+r \\cdot{ }^{(n-1)} P_{r-1}\\)
\n\u2234 \\({ }^{12} \\mathrm{P}_5+5 \\cdot{ }^{12} \\mathrm{P}_4={ }^{13} \\mathrm{P}_{\\mathrm{r}}\\)
\n\\({ }^{13} P_5={ }^{13} P_r\\)
\n\u2234 r = 5.<\/p>\n

Question 5.
\nIf \\({ }^{18} P_{(r-1)}{ }^{17} P_{(r-1)}\\) = 9 : 7, find r.
\nSolution:
\nGiven \\({ }^{18} P_{(r-1)}{ }^{17} P_{(r-1)}\\) = 9 : 7<\/p>\n

\"TS<\/p>\n

19 – r = 14
\nr = 5.<\/p>\n

Question 6.
\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 he in the same school.
\nSolution:
\nThe first son can be admitted to any one of the 5 schools \u00a1n 5 ways.
\nThe second son can be admitted to any one of the remaining 4 schools in 4 ways.
\nProceeding like this, the number of ways can he admit his sons in the schools so that no two of them will be in same school = 5 \u00d7 4 \u00d7 3 \u00d7 2 = \\({ }^5 \\mathrm{P}_4\\) (or) 120.<\/p>\n

\"TS<\/p>\n

II.
\nQuestion 1.
\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:
\nLet the ticket printed to enable a passenger to travel be from station ‘x’ to station ‘y’.
\nAs there are 25 railway stations on a railway line, ‘x’ can be filled in 25 ways and ‘y’ can be filled in remaining 24 ways.
\n\u2234 The required number of tickets = 25 \u00d7 24
\n= \\({ }^{25} \\mathrm{P}_2\\) or 600.<\/p>\n

Question 2.
\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:
\nNumber of students in a class = 30.
\nAs every student posts a greeting card to all his\/her classmates,
\nthe total number of greeting cards = \\({ }^{30} \\mathrm{P}_2\\) = 870.<\/p>\n

Question 3.
\nFind the number of ways of arranging the letters of the word TRIANGLE so that the relative positions of the vowels and conso-nants are not disturbed.
\nSolution:
\nThe given word TRIANGLE has 3 vowels and 5 consonants.
\nSince the relative positions of vowels and consonants are not to be disturbed.
\n3 vowels can be arranged in their relative positions in 3! ways.
\nSimilarly, 5 consonants can be arranged in their relative position in 5! ways.
\n\u2234 By fundamental principle of counting, the required number of ways = 3! \u00d7 5! = 720.<\/p>\n

\"TS<\/p>\n

Question 4.
\nFind the sum of all 4 digited numbers that can be formed using the digits 0, 2, 4, 7, 8, without repetition.
\nSolution:
\nGiven digits are 0, 2, 4, 7, 8.
\nFor a four digit number formed from given digits without repetition,
\nThousand’s place can be filled by non-zero digit in 4 ways,
\nHundred’s place can be filled by remaining 4 digits in 4 ways,
\nTen’s place can be filled in 3 ways and units place can be filled in 2 ways.
\n\u2234 By fundamental principle of counting, the number of four digited numbers formed = 4 \u00d7 4 \u00d7 3 \u00d7 2 = 96
\nOut of these 96 numbers,
\nthe numbers with ‘2’ in units place are 3 \u00d7 3 \u00d7 2 = 18
\nSimilarly the numbers with \u20192\u2019 in ten s place are 18.
\nThe numbers with ‘2’ in Hundred’s place are = 18.
\nThe numberlrwith ‘2’ in thousands place are \\({ }^4 \\mathrm{P}_3\\) = 24.
\n\u2234 The value obtained by adding ‘2’ in all the numbers = 18 (2) + 18 (20) + 18 (200) + 24 (2000)
\n= 18 \u00d7 2 (111) + 24 \u00d7 2 \u00d7 1000
\nSimilarly the value obtained by adding ‘4’ is 18 \u00d7 4 (111) + 24 \u00d7 4 \u00d7 1000
\nThe value obtained by adding ‘7’ is 18 (7) (111) + 24 \u00d7 7 \u00d7 1000
\nThe value obtained by adding ‘8’ is 18 \u00d7 8 \u00d7 111 + 24 \u00d7 8 \u00d7 1000
\n\u2234 The sum of all numbers = 18 \u00d7 2 \u00d7 111 + 24 \u00d7 2 \u00d7 1000 + 18 \u00d7 4 \u00d7 111 + 24 \u00d7 4 \u00d7 1000 + 18 \u00d7 7 \u00d7 111 + 24 \u00d7 7 \u00d7 1000 + 18 \u00d7 8 \u00d7 111 + 24 \u00d7 8 \u00d7 1000
\n= 18 \u00d7 111 \u00d7 (2 + 4 + 7 + 8) + 24 \u00d7 1000 \u00d7 (2 + 4 + 7 + 8)
\n= 18 \u00d7 111 \u00d7 21 + 24 \u00d7 1000 \u00d7 21
\n= 3 \u00d7 6 \u00d7 111 \u00d7 21 +4 \u00d7 6 \u00d7 1000 \u00d7 21
\n= 21 \u00d7 6 (333 + 4000)
\n= 21 \u00d7 6 (4333) = 5,45,958.<\/p>\n

Question 5.
\nFind the number of numbers that are greater than 4000 which can be formed using the digits 0, 2, 4, 6, 8 without repetition.
\nSolution:
\nGiven digits are 0, 2, 4, 6, 8.
\nAll the five digit numbers are greater than 4000.
\nIn the case of four digit numbers, the numbers which start with 4 or 6 or 8 are greater than 4000.
\nThe number of 4 digit numbers which starts with 4 or 6 or 8 using the given digits with out repetition are 3 \u00d7 \\({ }^4 P_3\\) = 3 \u00d7 24 = 72.
\nThe number of 5 digit numbers using the given digits with out repetition are 4 \u00d7 4! = 4 \u00d7 24 = 96
\n\u2234 The number of numbers which are greater than 4000 are 72 + 96 = 168.<\/p>\n

Question 6.
\nFind the number of ways of arranging the letters of the word MONDAY so that no vowel occupies even place.
\nSolution:
\nGiven word is MONDAY
\nGiven word contains 2 vowels and 4 conso-nants.
\nAlso given word contains 3 odd places and 3 even places.
\n. Since no vowel occupies even places, the two vowels in three odd places can be filled in \\({ }^3 \\mathrm{P}_2\\) ways.
\nThe four consonants can be filled in remain-ing four places in 4! ways.
\n\u2234 By fundamental principle of counting the required numbers of ways = \\({ }^3 \\mathrm{P}_2\\) \u00d7 4! = 144.<\/p>\n

\"TS<\/p>\n

Question 7.
\nFind the number of ways of arranging 5 different mathematics books, 4 different physics books and 3 different chemistry books such that the books of the same subject are together.
\nSolution:
\nGiven number of Mathematics books = 5
\nGiven number of Physics books = 4
\nGiven number of Chemistry books = 3
\nTreat all Mathematics books as 1 unit, Phys\u00acics books as 1 unit, Chemistry books as 1 unit.
\nThe number of ways of arranging 3 units of books = 3!
\n5 different Mathematics books can be arranged in 5! ways.
\n4 different Physics books can be arranged in 4!
\n3 different Chemistry books can be arranged in 3! ways.
\n\u2234. By fundamental principle of counting, required number of ways of arranging books = 3! \u00d7 5! \u00d7 4! \u00d7 3!
\n= 6 \u00d7 120 \u00d7 24 \u00d7 6 = 1,03,680.<\/p>\n

III.
\nQuestion 1.
\nFind the number of 5 letter words that can be formed using the letter of the word CONSIDER. How many of them begin with “C”, how many of them end with “R”, and how many of them begin with “C” and end with “R” ?
\nSolution:
\nGiven word “CONSIDER” contains 8 different letters.
\nThe number of 5 letter words that can be formed using the letters of word CONSIDER = \\({ }^8 \\mathrm{P}_5\\)
\nOut of them,
\ni) If first place is filled by ‘C’, the remaining 4 places by remaining 7 letters can be filled in \\({ }^7 \\mathrm{P}_4\\) ways.
\nNumber of words begin which ‘C’ are P4.
\nii) if last place is filled by \u2019R’, the remaining first four places by remaining 7 letters can be filled in \\({ }^7 \\mathrm{P}_4\\) ways.
\n\/. Number of words end with \u2019R’ are \\({ }^7 \\mathrm{P}_4\\).
\niii) If first place is filled with ‘C’ and last place is filled by ‘R’, the remaining 3 places between them by remaining 6 letters can be filled in \\({ }^6 \\mathrm{P}_3\\) ways.
\n\u2234 Number of words begin with ‘C’ and end with ‘R’ are \\({ }^6 \\mathrm{P}_3\\).<\/p>\n

Question 2.
\nFind the number of ways of seating 10 students A1<\/sub>, A2<\/sub>, ………….., A10<\/sub> in a row such that
\ni) A1<\/sub>, A2<\/sub>, A3<\/sub> sit together
\nii) A1<\/sub>, A2<\/sub>, A3<\/sub> sit in a specified order.
\niii) A1<\/sub>, A2<\/sub>, A3<\/sub> sit together in a specified order.
\nSolution:
\nA1<\/sub>, A2<\/sub>, ……………., A10<\/sub> are the 10 students.
\ni) A1<\/sub>, A2<\/sub>, A3<\/sub> sit together :
\nTreat A1<\/sub>, A2<\/sub>, A3<\/sub> as 1 unit.
\nThis unit with remaining 7 students can be arranged in 8! ways.
\nThe students A1<\/sub>, A2<\/sub>, A3<\/sub> can be arranged in 3! ways.
\n\u2234 The number of ways of seating 10 students such that A1<\/sub>, A2<\/sub>, A3<\/sub> sit together = 8! \u00d7 3!.<\/p>\n

ii) A1<\/sub>, A2<\/sub>, A3<\/sub> sit in a specified order :
\nIn 10 positions remaining 7 students other than A1<\/sub>, A2<\/sub>, A3<\/sub> can be arranged in \\({ }^{10} \\mathrm{P}_7\\)ways.
\nAs A1<\/sub>, A2<\/sub>, A3<\/sub> sit in a specified order, they can be arranged in 3 gaps in only 1 way.
\n\u2234 The number of ways of A1<\/sub>, A2<\/sub>, A2<\/sub> sit in a 10 specified order = \\({ }^{10} \\mathrm{P}_7\\).<\/p>\n

iii) A1<\/sub>, A2<\/sub>, A3<\/sub> sit together in a specified order :
\nTreat A1<\/sub>, A2<\/sub>, A3<\/sub> as 1 unit and they are in a specified order.
\nThis unit with remaining 7 students can be arranged in 8! ways.
\n\u2234 The number of ways of A1<\/sub>, A2<\/sub>, A3<\/sub> sit to-gether in specified order = 8!.<\/p>\n

\"TS<\/p>\n

Question 3.
\nFind the number of ways in which 5 red balls, 4 black balls of different sizes can be arranged in a row so that
\n(i) no two balls of the same colour come together
\n(ii) the balls of the same colour come together.
\nSolution:
\nGiven 5 red balls and 4 black balls are of different sizes.
\ni) No two balls of same colour come to-gether :
\nFirst arrange 4 black balls in row, which can be done in 4! ways \u00d7 B \u00d7 B \u00d7 B \u00d7 B \u00d7
\nThen we find 5 gaps, to arrange 5 red balls. This arrangement can be done in 5! ways.
\n\u2234 By principle of counting total number of ways of arranging = 5! \u00d7 4!.<\/p>\n

ii) The balls of same colour come together:
\nTreat all red balls as one unit and all black balls as another unit.
\nThe number of ways of arranging these two units = 2!
\nThe 5 red balls can be arranged in 5! ways while 4 black balls are arranged in 4! ways.
\nBy fundamental principal of counting, the required number of ways = 2! \u00d7 4! \u00d7 5!.<\/p>\n

Question 4.
\nFind the number of 4 – digit numbers that can be formed using the digits 1, 2, 5, 6, 7. How many of them are divisible by
\ni) 2
\nii) 3
\niii) 4
\niv) 5
\nv) 25
\nSolution:
\nThe number of 4 digit numbers that can be formed using the digits 1, 2, 5, 6, 7 with out repetition are \\({ }^5 \\mathrm{P}_4\\) or 120.<\/p>\n

i) Divisible by 2 :
\nA number with an even digit in its units place is divisible by ‘2’.
\nThis can be done in 2 ways (with 2 or 6). The remaining 3 places can be filled with the remaining 4 digits in \\({ }^4 P_3\\) ways.
\n\u2234 The number of 4 digit numbers, divisible by 2 are 2 \u00d7 \\({ }^4 P_3\\) = 48.<\/p>\n

ii) Divisible by 3 :
\nA number is divisible by 3 only when the sum of digits in that number is a multiple of 3.
\nSum of given 5 – digits is 21.
\nThe only way to select 4 digits using digits such that their sum is a multiple of 3 is selecting 1, 2, 5, 7.
\nWe can arrange then in 4! ways.
\n\u2234 The number of 4 digit numbers which are divisible by 3 are 4! = 24.<\/p>\n

iii) Divisible by 4 :
\nA number is divisible by 4 only when the number in last two places (i.e., ten’s and units) is a multiple of 4.
\n\u2234 The last two places should be filled with one of the following.
\n12, 16, 52, 56, 72, 76.
\ni.e., this can be done in 6 ways.
\nThe remaining 2 places can be filled with remaining 3 digits in \\({ }^3 \\mathrm{P}_2\\) ways.
\n\u2234 The number of 4 digit numbers which are divisible by 4 are 6 \u00d7 \\({ }^3 \\mathrm{P}_2\\)= 36.<\/p>\n

iv) Divisible by 5 :
\nA number is divisible by 5, if the units place is 0 or 5.
\nHence the units place from given digits is filled with ‘5’ only.
\nThe remaining 3 places by remaining 4 digits can be filled in \\({ }^4 \\mathrm{P}_3\\) ways.
\nThe number of 4 digit numbers which are divisible by 5 are \\({ }^4 \\mathrm{P}_3\\) = 24.<\/p>\n

v) Divisible by 25 :
\nA 4 digit number formed by using given digits is divisible by 25 if the number formed by ten’s and unit’s places is either 25 or 75. This can be done in 2 ways.
\nNow the remaining 2 places with remaining 3 digits can be filled in \\({ }^3 \\mathrm{P}_2\\) ways.
\n\u2234 The number of 4 digit numbers which are divisible by 25 are 2 \u00d7 \\({ }^3 \\mathrm{P}_2\\) = 12.<\/p>\n

\"TS<\/p>\n

Question 5.
\nIf the letters of the word MASTER are permuted in all possible ways and the words thus formed are arranged in the dictionary order, then find the ranks of the words
\ni) REMAST
\nii) MASTER.
\nSolution:
\nGiven word is MASTER.
\n\u2234 The alphabetical order of the letters is A, E, M, R, S, T.
\nIn the dictionary order, first we write all words beginning with ‘A’.
\nFilling first place with ‘A’, the remaining 5 places can be filled with the remaining 5 letters in 5! ways.
\n\u2234 The number of words begin with ‘A’ are 5! on proceeding like this, we get.<\/p>\n

i) Rank of the word REMAST :
\nThe number of words begin with A is 5! = 120
\nThe number of words begin with E is 5! = 120
\nThe number of words begin with M is 5! = 120
\nThe number of words begin with RA is 4! = 24
\nThe number of words begin with REA is 3! = 6
\nThe next word is REMAST.
\n\u2234 Rank of word REMAST = (120) 3 + 24 + 6 + 1 = 391.<\/p>\n

ii) Rank of the word MASTER :
\nThe number of words begin with A is 5! = 120
\nThe number of words begin with E is 5! = 120
\nThe number of words begin with MAE is 3! = 6
\nThe number of words begin with MAR is 3! = 6
\nThe number of words begin with MASE is 2! = 2
\nThe number of words begin with MASR is 2! = 2
\nThe next word is MASTER.
\n\u2234 Rank of word MASTER = 2 (120) + 2 (6) + 2 (2) + 1 = 257.<\/p>\n

Question 6.
\nIf the letters of the word BRING are per-muted in all possible ways and the words thus formed are arranged in the dictionary order, then find the 59th word.
\nSolution:
\nGiven word is BRING.
\n\u2234 The alphabetical order of the letters is B, G, I, N, R.
\nIn the dictionary order, first we write all words beginning with B.
\nClearly the number of words beginning with B are 4! = 24.
\nSimilarly the number of words begin with G are 4! = 24.
\nSince the words begin with B and G sum to 48, the 59th word must start with I.
\nNumber of words given with IB = 3! = 6
\nHence the 59th word must start with IG.
\nNumber of words begin with 1GB = 2! = 2
\nNumber of words begin with IGN = 2! = 2
\n\u2234 Next word is 59th = IGRBN.<\/p>\n

\"TS<\/p>\n

Question 7.
\nFind the sum of all 4 digited numbers that can be formed using the digits 1, 2, 4, 5, 6 without repetition.
\nSolution:
\nGiven digits are 1, 2, 4, 5, 6.
\nHence the number of four digit numbers formed using given digits with out repetition are \\({ }^5 P_4\\) = 120.
\nOut of these 120 numbers the numbers with 2 in unit’s place = \\({ }^4 P_3\\)
\nthe numbers with 2 in ten’s place = \\({ }^4 P_3\\)
\nthe numbers with 2 in Hundred’s place = \\({ }^4 P_3\\)
\nThe numbers with 2 in thousands place = \\({ }^4 P_3\\)
\n\u2234 The value obtained by adding ‘2’ in all-the numbers = \\({ }^4 P_3\\) \u00d7 2 + \\({ }^4 P_3\\) \u00d7 20 + \\({ }^4 P_3\\) \u00d7 200 + \\({ }^4 P_3\\) \u00d7 2000
\n= \\({ }^4 P_3\\) \u00d7 2 \u00d7 1111
\nSimilarly the value obtained by adding T in all the numbers is \\({ }^4 P_3\\) \u00d7 1 \u00d7 1111
\nThe value obtained by adding 4 in all the numbers is \\({ }^4 P_3\\) \u00d7 4 \u00d7 1111
\nThe value obtained by adding 5 in all the numbers is \\({ }^4 P_3\\) \u00d7 5 \u00d7 1111
\nThe value obtained by adding 6 in all the numbers is \\({ }^4 P_3\\) \u00d7 6 \u00d7 1111
\n.\\ The sum of all the numbers = \\({ }^4 P_3\\) \u00d7 1 \u00d7 1111 + \\({ }^4 P_3\\) \u00d7 2 \u00d7 1111 + \\({ }^4 P_3\\) \u00d7 4 \u00d7 1111 + \\({ }^4 P_3\\) \u00d7 5 \u00d7 1111 + \\({ }^4 P_3\\) \u00d7 6 \u00d7 1111
\n= \\({ }^4 P_3\\) \u00d7 (1111) (1 + 2 + 4 + 5 + 6)
\n= 24 \u00d7 1111 \u00d7 18 = 4,79,952
\nThe sum of all 4 digit numbers that can be formed with given digits is 4,79, 952.<\/p>\n

Question 8.
\nThere are 9 objects and 9 boxes. Out of 9 objects, 5 cannot fit in three small boxes. How many arrangements can be made such that each object can be put in one box only.
\nSolution:
\nGiven that there are 9 objects and 9 boxes.
\nAs 5 objects out of 9, cannot fit in 3 small boxes, these 5 objects should be arranged in . remaining 6 boxes.
\nThis can be done in \\({ }^6 \\mathrm{P}_5\\) ways.
\n\u2234 The remaining 4 blanks are to be filled with remaining 4 objects.
\nThis can be done in 4! ways.
\n\u2234 The required number of ways = \\({ }^6 \\mathrm{P}_5\\) \u00d7 4! = 17,280.<\/p>\n","protected":false},"excerpt":{"rendered":"

Students must practice this TS Intermediate Maths 2A Solutions Chapter 5 Permutations and Combinations Ex 5(a) to find a better approach to solving the problems. TS Inter 2nd Year Maths 2A Solutions Chapter 5 Permutations and Combinations Ex 5(a) I. Question 1. If = 1320, find n. Solution: Given = 1320 \u21d2 n(n – 1) … Read more<\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/9422"}],"collection":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/comments?post=9422"}],"version-history":[{"count":2,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/9422\/revisions"}],"predecessor-version":[{"id":9426,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/9422\/revisions\/9426"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=9422"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=9422"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=9422"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}