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How many numers are there between 100 and 1000 such that

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New post 25 Jan 2004, 16:05
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How many numers are there between 100 and 1000 such that atleast one of their digits is 4?

648
652
252
248
387
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New post 26 Jan 2004, 10:01
Ans: 252

Lets assume xxx are the 3 positions for digits,

If 4 is at the hundereds poisiton, then I can 10 different options for tenth position and 10 different option for unit position, which will give 10*10=100

Now say 4 is at Tens position, which will give 8 options for hunderds position(excluding 0 and 4, as 4 at hundereds is covered above) and 10 options for unit place, 8*10= 80

4 at units place, will give 8 options for hundereds place and 9 options for tens place(excluding 4 at tens place as it is covered above), which will give 8*9= 72

Hence total ways is 100+80+72= 252.
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New post 26 Jan 2004, 11:30
An alternative approach that may end up being slightly shorter timewise:

lets count ALL the numbers and then SUBTRACT those that DO NOT include a 4. (I understand between 100 and 1000 as meaning - from 101 to 999 but apparently the meaning here is from 100 to 1000).

We ignore 1000 as we don't need to count it anyway

Then we are left with all 3 digit numbers. There are 8 options for the first (hundreds digit - as it cannot be 0 or else we would get a 2 digit #, or 4 because we don't allow 4). There are 9 options for the tens and units (because we dont allow 4). so there are 8x9x9 = 648 number that do not have a 4 in them.

total numbers equal - 999-100+1 = 900.

900 - 648 = 252.

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New post 26 Jan 2004, 13:58
:cool
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Pls include reasoning along with all answer posts.
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  [#permalink] 26 Jan 2004, 13:58
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