[PS][arragements] HSPA posts

(Assuming no repetition)

IMO it should be C (125)
two digit numbers can be formed in 5*5=25 ways
three digit numbers can be formed in 5*5*4 = 100 ways
total = 125 numbers

IF REPETITION ALLOWED:
2-digit numbers: 5*6=30 numbers
3 digit numbers: 5*6*6 = 180 numbers
total = 210 numbers
Since this is not in answer choices, i assume no repetition was allowed

assuming repetition is not allowed

two digit numbers 5*5 =25 ways
three digit numbers 5*5*4 =100 ways
hence total numbers possible = 125 numbers

Hence C

Guys,
help me. I m i reading too much into the Q?
it asks for count of numbers btw 10 and 1000 having digits 2,3,4,0,8,9
so we are only looking for numbers wth these digits or can we take 21 to be part of this list?
21- it does contain 2(one of the given numbers)
we would get more combinations then.

rohu27 wrote:Guys,
help me. I m i reading too much into the Q?
it asks for count of numbers btw 10 and 1000 having digits 2,3,4,0,8,9
so we are only looking for numbers wth these digits or can we take 21 to be part of this list?
21- it does contain 2(one of the given numbers)
we would get more combinations then.

here we have to form the numbers between 10 and 1000 by using the digits (2,3,4,0,8,9)..!! since 1 is not the part of the given digits so 1 won't be used i.e. combinations like 21,31,13,12 etc. are not legal... as per the question..!!!

Good job guys.. OA is pointing 125