A box contains 20 tickets of identical appearance, the ticket being numbered 1, 2 , 3.....20. In how many ways can 3 tickets be chosen such that the numbers on the drawn tickets are in arithmetic progression?If sumone cud post the approach it wud be gr8

1) 18
2) 33
3) 56
4) 90
5) 84

• Ans is 90
  there could be maximum Common difference will be 9 so we have to check no of ways of AP with Common difference form 1 to 9
  
  with C.D. 1 : 18 ways
  with C.D. 1 : 16 ways
  with C.D. 1 : 14 ways
  with C.D. 1 : 12 ways
  with C.D. 1 : 10 ways
  with C.D. 1 : 8 ways
  with C.D. 1 : 6 ways
  with C.D. 1 : 4 ways
  with C.D. 1 : 2 ways
  
  TOtal No. of ways are 2 + 4 + 6 + 8 + 10 + 12 + 14 +16 + 18
  with is 90
• 9 years agoHelpfull: Yes(5) No(4)
• 2+4+6+..+18=90
• 10 years agoHelpfull: Yes(4) No(6)
• 10c2*2!=90
• 10 years agoHelpfull: Yes(4) No(4)
• A sequence of 3 tickets, starting at "a" and with a difference d, gives
  a, (d + a), (2d + a)
  
  For a given d, put a = 1, then final ticket number is
  2d + 1
  
  There are 20 - (2d + 1) ticket numbers above this final ticket
  Thus, keeping the same d, we can increment the start point
  20 - (2d + 1) times and stay in range
  [ie slide the whole sequence along]
  
  So for any given d there are
  1 + 20 - (2d +1)
  = 20 - 2d
  = 2(10 - d)
  possible ways to make an AP, and d = 1 to 9
  
  So adding up for all d, we just get twice the sum of 1 to 9
  ie sum of an AP
  
  2{ (1/2)(9 x 10) } = 90
• 10 years agoHelpfull: Yes(3) No(6)
• consider intial a,b,c as progession
  a=1,then b=x,c=2x-1.....beacuse c=a/2+b/2...then c is odd ...max of c=19,then x=10........min to max values of for a=1 are 2 to 10......9 ways......
  a=2, x varies from 3 to 11....ways = 11-3+1=9..
  a=3,ways =4 to 11...ways=8
  a=4, ways =8.......
  u will a similarity such that max a=18..
  similarity= a=1,2...ways =9+9=9*2
  a=2,3..ways=8*2.......son on
  at a=17,18..ways=1+1=1*2
  ..............total no of ways=2*(1+2+3+4+5+6+7+8+9)=45*2=90
  option D=90
• 7 years agoHelpfull: Yes(0) No(2)
• Common Difference(CD)-1 : (1,2,3),(2,3,4),(3,4,5),.............,(1... sets
  CD-2 : (1,3,5),(2,4,6),(3,5,7),.........,(16,18... sets
  CD-3 : (1,4,7),(2,5,8),(3,6,9),........,(14,17,... sets
  and so on upto CD-8
  for CD-9 there are only two sets i.e., (1,10,19),(2,11,20)
  therefore,sum of all sets =18+16+14+............+4+2=90
• 6 years agoHelpfull: Yes(0) No(0)