1,2,3,4,5,6,7 are arranged such that sum of two successive numbers is a prime number....ex-1234765(i.e. 1+2=3, 2+3=5, 3+4=7....)

Q1-how many such possible combinations occur?
Q-2-How many possible combination occurs if first number is 1/7 and last number is 7/1(i.e 1xxxxx7 or 7xxxxx1)?
Q-3-How many numbers will come on 4th position(xxx_xxx)?

• Total possible numbers are:
  1234765
  3214765
  5234761
  5234167
  1674325
  5674321
  7614325
  1476523
  3476523
  7416523
  1256743
  3256743
  5216743
  1652347
  5612347
  7652143
  7652341
  1432567
  3412567
  7432165
  7432561
  These numbers is obtained with the help of hit & trial in
  (_2_4_6_), (_6_4_2_), (_4_6_2_), (_2_6_4_), (_6_2_4_), (_4_2_6_)
  Since question uses logic of oeoeoeo.
  Q1.) 21
  Q2.) 4
  Q3.) 3
• 8 years agoHelpfull: Yes(7) No(1)
• q2 ans is 4
  7432561
  7652341
  1652347
  1432567
• 9 years agoHelpfull: Yes(6) No(1)
• I am appreciate to all those who give ideas to treat the problem with regards of odd and even number and their positions. Accordingly the format of the combination will be- "oeoeoeo" where 'o' stands for odd number and 'e' stands for even number.
  
  But, the thing should be notice here is - "2 cun't stay with 7" and also "4 cun't stay with 5" as in both the cases the sum result in 9, which is not a prime number!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
• 9 years agoHelpfull: Yes(3) No(0)
• Following by my previous post (see above/below) the solution should be such that
  a) Total number of arrangement by cosidering the order 'oeoeoeo'
  b) Then by removing the terms which occurs as 'succisive 2 and 7' also 'succisive 4 and 5' seperately
  c) Then by adding the number of terms those consists of both 'succisive 2 and 7' and 'succisive 4 and 5' because in the last step (b) this types of terms (the intersection) was removed two times instead of one time.
  
  Now executing_ (a)
  all the 4 'o' can be permute as 4!
  all the 3 'e' can be permute as 3!
  hence a = 4!*3!
  
  Now executing_ (b)
  first consider the case of 2 followed by 7 {
  It can take any of 3 availabe position of 'eo' it can be choosen by 3C1
  Rest of_
  all the 3 'o' can be permute as 3!
  all the 2 'e' can be permute as 2!
  hence such types os cases = 3C1*3!*2!
  }
  now 7 can also follow 2
  so, total number of cases in 'succisive 2 and 7' = 2*3C1*3!*2!
  similar cases will occus while considering 'succisive 4 and 5'
  so, total number of cases under (b)
  b = 2*2*3C1*3!*2!
  
  Now exexuting_ (c)
  first consider even followed by odd ie. 27 and 45 {
  out of 3 'oe' 2 can be filled by 27, 45 and can permute among themselv by 3P2
  Rest of_
  Both the 2 'o' can be permute as 2!
  1 'e' ie. 6 will take the remaining only one vacant position
  hence such types os cases = 3P2*2!
  }
  similar cases will occus when odd is followed by even ie. 72 and 54
  hence total number of cases either odd followed by even/ even followed by odd = 2*3P2*2!
  now the cases left are_
  oeoeoeo
  c1_ 72o45eo with 2! ways
  c2_ 72oeo45 with 2! ways
  c3_ oe72o45 with 2! ways
  c4_ o45e72o with 2! ways
  and also
  c5_ 54o27eo with 2! ways
  c6_ 54oeo27 with 2! ways
  c7_ oe54o27 with 2! ways
  c8_ o27e54o with 2! ways
  gives total 8*2! cases. (2! ways because only 2 'o' are able to permute in 2! ways)
  so. total number of cases under (c)
  c = 2*3P2*2! + 8*2!
  
  
  
  Hence result of Q1. is
  a-b+c = 4!*3! - 2*2*3C1*3!*2! + (2*3P2*2! + 8*2!)
  = 40 numbers of possible combination
  
  I won't repeat the result of Q2, it's correct that answered by Mr. 'AS'
  and along with everyone my answer Q3 goes for 3.
• 9 years agoHelpfull: Yes(3) No(0)
• Q1.
  1674325
  5674123
  3214765
  7614325
  5234167
  5674321
  1234765
  Q2. Combinations of 1......7/ 7........1:
  0
  Q3.
  1
  
  
  
  
• 9 years agoHelpfull: Yes(1) No(3)
• see even no. will occupy even position and odd no. will occupu odd position(chk urself)
  keeping 4th element 2,4,6 each will give 2 combination and reversing that will give 2 more combination.
  (a)4*3=12 combination (m not sure can be more )
  (b)2 combinations are posible keeping no.2 at 4th position (1432567 and reversing)
  (c)3(as there are 3 even no.)
• 9 years agoHelpfull: Yes(0) No(2)
• the 2nd answer is 3 as because the no are 1652347, 7432561,7652341
• 9 years agoHelpfull: Yes(0) No(1)
• we...know..e+o=o....,annd here ..total..odd no.=4,and..even ..no.=3..thus..total possible arrengement..willbe.........4*3*3*2*2*1*1=144
• 9 years agoHelpfull: Yes(0) No(1)
• @ SANJOY DEBNATH; Excellent approach , but i not understand why are you ignoring the case of (3,6) & 6,3) ?
• 8 years agoHelpfull: Yes(0) No(0)