Zero has not been considered above. So when you consider zero, then one of the other number should be twice of another. there are 4 possibilities and each has 4 ways..so adding 16 the answer is 96+16=112. Answer choice B)
the case for numbers like 111,222,333.........999 has not been considered... 9 cases more are added thus my answer appears to be(96+16+9)=121 would anyone like to correct me?
Yeah @wolverine your solution counts for those aps when terms are equal but this is an old amc problem and they did not give option for counting with that approach
so the option 121 is not there answer is right
so as the term distinct is not mentioned we should also count for there 9!
96
ReplyDelete(1,2,3),(2,3,4)....(7,8,9) 7 triplets
(1,3,5),(2,4,6)...(5,7,9) 5 triplets
(1,4,7),(2,5,8),(3,6,9) 3 triplets
(1,5,8) 1 triplet
Total possible triplets =16
Total possible no.s =16* 3! =96
Zero has not been considered above. So when you consider zero, then one of the other number should be twice of another. there are 4 possibilities and each has 4 ways..so adding 16 the answer is 96+16=112. Answer choice B)
ReplyDeletenice way...
ReplyDelete(a+b)/2=c
consider the terms in ap and arrange :)
@Celebrating Life
ReplyDeletethe case for numbers like 111,222,333.........999
has not been considered...
9 cases more are added thus my answer appears to be(96+16+9)=121
would anyone like to correct me?
Yeah @wolverine
ReplyDeleteyour solution counts for those aps when terms are equal
but this is an old amc problem
and they did not give option for counting with that approach
so the option 121 is not there
answer is right
so as the term distinct is not mentioned
we should also count for there 9!
Good Job
then i must say it was a flawed question...!!!
ReplyDeleteyupp, it was ambiguity is generally avoided by amc
ReplyDelete16*3!+9+4*(3!-2!)=121
ReplyDelete