1,2,3,4,5,6,7 are arranged such that sum of two successive numbers is a prime number. For example, 1234765 (i.e. 1+2=3, 2+3=5, 3+4=7....)
1. How many such possible combinations occur?
2. How many possible combination occurs if first number is 1/7 and last number is 7/1 (i.e 1xxxxx7 or 7xxxxx1)?
3. How many numbers will come on 4th position(xxx_xxx)?

• We have to follow a systematic approach for this question. We know that if at all two consecutive numbers be prime, they should not be even in the first place. So we arrange even numbers in even places, odd numbers in odd places.
  2 should not have 7 as its neighbor
  4 should not have 5 as its neighbor
  6 should not have 3 as its neighbor
  Case 1:
  __ 2 ____ 4 ____ 6 __
  x7 x7,5 x5,3 x3
  So fix, 3 or 1 in between 2 and 4. We get the following options with reverse case also.
  5234167, 7614325
  5234761, 1674325
  1234765, 5674321
  3214765, 5674123
  
  Case 2:
  __ 2 ____ 6 ____ 4 __
  x7 x7,3 x3,5 x5
  
  5216743, 3476125
  3256147, 7416523
  3256741, 1476523
  1256743, 3476521
  
  Case 3:
  __ 4 ____ 2 ____ 6 __
  x5 x5,7 x7,3 x3
  
  3412567, 7652143
  7432165, 5612347
  7432561, 1652347
  1432567, 7652341
  Answer 1: Total possibilities are 24.
  Answer 2: 4 possibilities.
  Answer 3: 3 possibilities.
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