How many three-digit numbers can be formed from the given set if each digit can occur only once? Of these, how many are even

combinatoricspermutations

A) How many three digit numbers from $1, 2, 3, 4, 5, 6, 7, 8, 9$ can be created?

I answered this by $9 \times 8 \times 7 = 504$. The possible answers are 84 – 450 – 504 – 729

B) How many of those numbers are odd?

Possible answers are 42 – 405 – 280 – 225

I tried answering this with $9 \times 8 \times 5$ (since it should end with an odd number and there are $5$ of those in the offered numbers). However, it is not correct.

EDIT: I forgot to add that every number can only occur once.

Suggestions? Thank you in advance!

Best Answer

If every number can be used once , you must select first the unit cipher and you have $5$ choices, then you have only $8$ choices and then $7$, so the answer is:

$$5\times 8 \times 7=280$$

:)