Posted by **Andrew** on Wednesday, November 14, 2012 at 11:12pm.

if only digits 0,1,2,3,4,5,6, and 7 may be used find number of possibilites four digit numbers

- math -
**Alexa**, Wednesday, November 14, 2012 at 11:17pm
n!/r!(n-r)!

8!/4!(8-4)!

= 70

- math -
**Reiny**, Wednesday, November 14, 2012 at 11:18pm
Ok, Andrew

in my last answer to the same question I somehow missed the 7 and saw only 6 numbers.

(Have to concentrate more on reading the question more carefully)

In each case we can assume that no whole number starts with a zero, and no repeating digits for all 3 cases.

We have 8 digits, but can only use 7 in the first position ...

number of 4 digits numbers = 7x7x6x5 = 1470

number of 3 digit odd numbers, (must end in an odd) = 6x6x4 = 144

number of 3 digit numbers = 7x7x6 = 294

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