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counting and probability

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by kvcpk » Tue Jun 15, 2010 2:16 am
If first digit is 9, there can be 100 numbers (10*10 )
If second digit is 9, there can be 100 numbers (10*9 ) [9 for first digit bcos it cant be zero]
If third digit is 9, there can be 100 numbers (10*9 ) [9 for first digit bcos it cant be zero]

so total 100+90+90 = 280 numbers

OA?
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by sumanr84 » Tue Jun 15, 2010 5:37 am
duongthang wrote:find the number of 3-digit numbers abc that has exactly 1 digit 9 in it.
Notice the catch: exactly 1 digit 9..So on a similar note as kvcpk answer should be 225

9 X X => 9 * 9 = 81

X 9 X => 8 * 9 = 72

X X 9 => 8 * 9 = 72

Total = 225
I am on a break !!
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by kvcpk » Tue Jun 15, 2010 7:40 am
sumanr84 wrote:
duongthang wrote:find the number of 3-digit numbers abc that has exactly 1 digit 9 in it.
Notice the catch: exactly 1 digit 9..So on a similar note as kvcpk answer should be 225

9 X X => 9 * 9 = 81

X 9 X => 8 * 9 = 72

X X 9 => 8 * 9 = 72

Total = 225
Thanks Suman for the correction. I missed it.
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