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?”
On
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:
$$3 + 2 \cdot 3 = 9$$
On
You have two criteria.
Just count the ways this works.
On
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
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: $$412, 432, 452, 512, 514, 524, 532, 534, 542.$$