How many even numbers over $400$ can be made out of the integer set $2$, $4$, and $7$ if each integer is used only once?
That's the question. My answer is,
There are $3$ digits. To be an even number the last digit should be $2$ or $4$. So $2$ possibilities. Then I'm left out with another $2$ digits. So,
$$2 \times 1 \times 2 = 4$$
But in reality I can only see $3$ numbers which can be made from the given integers.
- $472$
- $742$
- $724$
So what's the proper way to solve this? Please show me the way to get the answer $3$.
You have correctly identified that the last digit needs to be even. However, you have failed to take into account that the first digit needs to be greater than or equal to $4$. The way the numbers work out in this particular problem makes this not change the equation, but you jumped to an okay equation (though as you noted not perfect) without a complete argument for it. $2\times 1\times2$ is the formula for the number of ways to make a number starting with a number >= 4 and ending with an even number, except....
The other issue is that you are failing to take into account dependent probability. If the first number is a $4$, then there is only one possible answer, $472$. You can’t make a number starting in $4$ and ending in $4$ because you can only use $4$ once.
Try using inclusion-exclusion.