A paradox in counting number of 3-digit odd integers having distinct digits

Consider the following two methods for counting number of 3-digit odd integers $\overline{a_2a_1a_0}$ having distinct digits:
Method 1)
we have 5 choices for $a_0$ , 10-1 choices for $a_1$ and 9-1-1 choices for $a_2$ which gives a total count of: $5\times9\times7=315$
Method 2)
we have 5 choices for $a_0$ , 9-1 choices for $a_2$ and 10-1-1 choices for $a_1$ which gives a total count of $5\times8\times8=320$.
We do know the the correct answer is 320,so what's wrong with first method?

There are $9-1-1$ choices for $a_2$ IF you've chosen $a_1\not=0$. If $a_1=0$, then there are $10-1-1$ choices instead.

• So can you correct the first method to give correct answer??? Sep 29 '15 at 7:12
• The first method is using the multiplication rule incorrectly. The only way to correct it would be to consider whether $a_2=0$ by cases. If so, then there are $5\cdot 1\cdot 8$ odd numbers; if not, then there are $5\cdot 8\cdot 7$ numbers. $5\cdot1\cdot8+5\cdot8\cdot7=40+280=320$ total. Sep 29 '15 at 7:14
• Excellent! Thank you,but I think you have a mistake in your last comment mentioning $a_2$ instead of $a_1$, don't you? Sep 29 '15 at 7:16
• Yeah. It's too late to fix it, though. Sep 29 '15 at 7:22
• Could you please explain why should we consider 2 cases for $a_1$ in first method? Why not doing that makes first method wrong???? Sep 29 '15 at 7:24

I think it is more clear if you consider the order you select. I mean the following :

1. First select $a_0$. It is clear there are 5 choices, namely (1,3,5,7,9)
2. The consider choices for $a_2$. It cannot be $0$ (not 3 digit number) and it cannot be $a_0$ (distinct condition). Thus there are $8$ choices.
3. Finally there are also $8$ choices for $a_1$, since it cannot be $=a_2$ neither $=a_0$.

So the correct answer is $5 \times 8 \times 8 = 320$

• Our main problem is with first method: If we decide to check $a_1$ before $a_2$,why should we distinguish two cases for $a_1$??? Sep 29 '15 at 8:11
• I don't see two cases for $a_1$. Simply if you have to find all possible choices (for $a_1$) then you have to look at the whole picture. Put it in other words, it is not a paradox, but if you follow the first method it is easy to get tricked. Sep 29 '15 at 8:19
• Finally I figured it out: If we check possible cases of $a_1$ before $a_2$ then we must consider that when $a_1=0$ there are 8 possible choices for $a_2$ and when $a_1 \neq 0$ there are 7 possible choices for $a_2$! That's all. Sep 29 '15 at 12:53

Finally I figured it out:
If we check possible cases of $a_1$ before $a_2$ then we $must$ consider that when $a_1=0$ there are 8 possible choices for $a_2$ and when $a_1\neq0$ there are 7 possible choices for $a_2$! That's all.