How many ways to build a 3-digit odd number (no repetition)

Question

I searched and could only find an answer for the problem with repetition of the following question.

How many 3-digit odd numbers greater than 600 can be created using the digits 2,3,4,5,6, and 7? Repetition of digits is not allowed.

I found the correct answer, $20$, by creating a tree, but I am wondering if there is a more elegant way of counting this.

Answer 1

If you don't worry whether the number is even or odd, just that it's bigger than $600$, then the count is $2\cdot5\cdot4=40$, since the first digit must be $6$ or $7$ and the other two can be anything (that's not already been used). Since there are the same number of even and odd digits (three of each), and also the same number available for the lead digit (one of each), the result is symmetric between even and odd, so the number of each is $40/2=20$.

Remark: This only works as nicely as it does because of the symmetry (i.e., equality) in the counts of even and odd digits. If you were using the digits $1$ to $7$, say, or $2$ to $8$ it wouldn't work, and you'd have to go with something case-based.

Answer 2

Interesting question:

How many ways to build a $3$-digit odd number (no repetition)

We have $5$ odd numbers, namely, $1, 3, 5, 7, 9$ and we also have $3$ spots where we could put the numbers. __ __ __ In the first one we could place any of the 5 numbers that we have, then $4$ is left and finally $3$ so it would be $5*4*3 = 5P3 = 60$ odd numbers.

If repetition is allowed, it would be $5^3$

That could be done in $5\choose3$ to choose where we will place them and then multiply by 3$!$ to arrange them among themselves ways which would give 1$0*6 = 60$ again. Depends on if the order is important or not.

How many $3$-digit odd numbers greater than $600$ can be created using the digits $2,3,4,5,6,$ and $7$? Repetition of digits is not allowed.

If we are only to use odd numbers, and we want the number to be greater than 600, then we should start by $6$, or $7$ which would be $2$ possibilities, then we have 4 remaining digits out of the 5, let's choose now any of the 5 digits, and then 4 would be left so it would give $2*5*4 = 40$ and don't forget about the leading digits which would divide the answer by 2.