Thread: Counting Principle

How many ways to form a 6-digit number which the digits cannot be repeated and the leftmost digit cannot be 0?

I found there're two possible answer:
since 0 is not available to the leftmost digit, so there're only 9 choices and 9 choices on the second digit(since 0 is available) in the second digit, 8 choices on third digit, and so on. The total number of way is: 9*9*8*7*6*5 = 136080

but if I start at the rightmost digit, the answer will be different!!!
There are 10 choices at the rightmost digit, 9 choice at the second-rightmost digit, and so on. There are 6 choices at second-leftmost digit, 4 choices left at the leftmost digit (since 0 is unavailable here). The total number of ways is: 10*9*8*7*6*4 = 120960.

Could anyone explain this?

Originally Posted by piscoau

How many ways to form a 6-digit number which the digits cannot be repeated and the leftmost digit cannot be 0?
I found there're two possible answer:
since 0 is not available to the leftmost digit, so there're only 9 choices and 9 choices on the second digit(since 0 is available) in the second digit, 8 choices on third digit, and so on. The total number of way is: 9*9*8*7*6*5 = 13608f

but if I start at the rightmost digit, the answer will be different!!

You should not get different answers.
If you notice
That is starting on the right. The first product counts no zero at all, the second counts one zero in the first five digits right to left.