Question 1055418
How many four-digit odd numbers can be formed from the digits 
1, 2, 4, 5, 6, and 9 if each digit can only be used once?
<pre><b>
Rule for this type problem:

<i>Choose the most restrictive thing(s) first!</I>

Since the number must be odd, the last digit must
be odd.

So the last digit is the most restrictive thing to choose.

So we choose the last digit first:

We can choose the last, or 4th, digit 3 ways, 1, 5, or 9.
We can then choose the 1st digit 5 ways
(as any of the remaining 5 digits).
We can then choose the 2nd digit 4 ways
(as any of the remaining 4 digits).
We can then choose the 3rd digit 3 ways
(as any of the remaining 3 digits).

That's 3×5×4×3 = 180 ways.

Edwin</pre><b>