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How many odd three-digit integers....

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by tanyasethi » Tue Jul 12, 2011 1:52 pm
How many odd three-digit integers greater than 800 are there such that all their digits are different?
  • 40
    56
    72
    81
    104
The correct answer is C, but I don't understand how.
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by Ian Stewart » Tue Jul 12, 2011 3:48 pm
If our number is between 800 and 899, we have 5 choices for the units digit, since it must be odd. Now, the tens digit cannot be the same as the units digit and also cannot be the same as the hundreds digit, so there are 8 remaining choices for the tens digit. There are thus 5*8 = 40 odd numbers with three different digits between 800 and 899.

If our number is between 900 and 999, we only have 4 choices for the units digit, since it cannot be 9. We again have 8 choices for the tens digit, so there are 4*8 = 32 odd numbers with three different digits between 900 and 999.

Adding the results from each case, we have 40+32 = 72 numbers in total.
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by amit2k9 » Tue Jul 12, 2011 11:37 pm
between 800-899
hundred digit = 1 choice
tens digit =9 choices (include 0)
ones digit = 8 choices
thus total = 9*8=72 odd digits = 72/2=36. (using symmetry)

similarly 72/2 = 36 odd digits for integers between 900-999.

thus total 36*2 = 72.
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