Question 1202146: How many 8 digit numbers can be formed if the leading cannot be a zero and the last number cannot be 1
Answer by ikleyn(52814) (Show Source):
You can put this solution on YOUR website! .
If repeating is allowed (which I can suppose from the context),
then the answer is 9*10*10*10*10*10*10*9 = 81,000,000.
Of 8 multipliers, two multipliers of 9 correspond to the first and the last positions, where restrictions are imposed;
six multipliers of 10 correspond to six other positions with no restrictions.
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