# (Gr. 10) How many $3$-digit even numbers greater than $400$ can be formed...

Please help! The answer in our textbook is $$2$$, but I assume that’s a misprint. I answered it myself and I got $$12$$ possible numbers $$(1 \cdot 3 \cdot 1) + (1 \cdot 2 \cdot 2)$$ but I just want to make sure :)

“How many $$3$$ digit even numbers greater than $$400$$ can be formed from the digits $$\{1, 2, 3, 4, 5\}$$ if repetition of digits is not allowed?”

• Hey and Welcome to MSE! Would you care explaining a bit more how did you come up with your solution? Feb 25, 2019 at 12:00
• I am also pretty sure that it is a misprint, $2$ seems terribly few Feb 25, 2019 at 12:01
• I am sure that your second parenthese is incorrect, how did you get this number? Feb 25, 2019 at 12:12

If the number starts with a $$4$$, it must end with a $$2$$. Three options for the second digit remain.

If the number starts with a $$5$$, it can end with a $$2$$ or a $$4$$. Again, three options for the second digit remain.

The total number of valid numbers thus equals:

$$3 + 2 \cdot 3 = 9$$

The answer is $$9$$. The number must start with a $$4$$ or a $$5$$ and must end with a $$2$$ or a $$4$$, so the possibilities are: $$412, 432, 452, 512, 514, 524, 532, 534, 542.$$

You have two criteria.

1. The end digit must be $$2$$ or $$4$$.
2. The first digit must be $$4$$ or $$5$$.

Just count the ways this works.

You are to create $$3$$ digit even numbers greater than $$400$$ so think as follows:

It cannot start with a number less than $$4$$. So let us start with $$4$$, then you only have one choice left for the last digit, namely $$2$$. Can you figure out the rest? (you can peek by mouseovering the yellow boxes, but try to figure out yourself first, before you peek)

$$3$$ different numbers since the first and last digit is determined and you have $$3$$ other digits to choose from to fill the middle

It can also start with a $$5$$, in this case it can end with both $$2$$ and a $$4$$. I am sure you can figure the rest out ;)

If it ends with a $$2$$ you have $$3$$ choices for the middle number, since once again we fixed the first and the last digit and the same is true for when it ends with a $$4$$

Adding these together you end up with

a total number of $$3+3+3=9$$ possibilities.

Hope I could help