3 numbers have to be choosed from 1 to 9 in such a way they always becomes an A.P series. Find the probability?please give the explaination.??

• (1,2,3),(2,3,4),(3,4,5),(4,5,6),(5,6,7),(6,7,8),(7,8,9)-diff-1= 7
  (1,3,5),(2,4,6),(3,5,7),(4,6,8),(5,7,9)-diff-2= 5
  (1,4,7),(2,5,8),(3,6,9)-diff-3= 3
  (1,5,9)-diff-4= 1
  so total series will be (7+5+3+1)*2=32 as they can be arranged in reverse order
  and the numbers can be chosen by 9C3 ways
  so the ans is 32/9C3
  
• 10 years agoHelpfull: Yes(63) No(5)
• 3 no. can be chosen from 9 in 9C3 ways.
  For the no. for being in a.p. The value 0f d will be {1,2,3,4,}
  d=common difference
  for d =1 there is 7 ways to chose from
  for d=2 there is 5 ways to chose from
  for d=3 there is 3 ways
  for d=4 there is 1 way
  so 16/(9C3)
• 10 years agoHelpfull: Yes(7) No(6)
• there are 32 outcom....
  and hence probability = 32/9C3.
• 10 years agoHelpfull: Yes(5) No(0)
• 3 no. can be chosen from 9 in 9P3 ways.
  (here we r considering reverse order so order of the numbers is important)
  for a.p. The value of d will be {1,2,3,4,}
  for d =1 there is 7 ways
  for d=2 there is 5 ways
  for d=3 there is 3 ways
  for d=4 there is 1 way
  now (16*2) 32 ways cosidering negative values of d
  ans 32/9p3
• 10 years agoHelpfull: Yes(4) No(5)
• it should be 16*2/9C3 which is 8/21
• 10 years agoHelpfull: Yes(3) No(0)
• 3 no. can be chosen from 9 in 9C3 ways.
  For the no. for being in a.p. The value 0f d will be {1,2,3,4,}
  d=common difference
  for d =1 there is 7 ways to chose from
  for d=2 there is 5 ways to chose from
  for d=3 there is 3 ways
  for d=4 there is 1 way
  so 9/(9C3)
• 10 years agoHelpfull: Yes(2) No(13)