# Slips of paper numbers 1 through 14 are placed in a hat. In how many ways can you draw two numbers with replacement that total 12?

Jun 11, 2018

$11$ ways

#### Explanation:

Say that your first draw is $x$ and the second draw is $y$. If you want $x + y = 12$, you can't have $x = 12 , 13 \mathmr{and} 14$. In fact, since $y$ is at least one, $x + y \setminus \ge x + 1 > x$

So, assume that the first draw is $x \setminus \in \setminus \left\{1 , 2 , \ldots , 11 \setminus\right\}$. How many "good" values for $y$ we have for each of these draws?

Well, if $x = 1$, we must draw $y = 11$ in order to have $x + y = 12$. If $x = 2$, $y$ must be $10$, and so on. Since we allow replacement, we can include the case $x = y = 6$ as well.

So, we have $11$ possible values for $x$, each yielding exactly one value for $y$ in order to have $x + y = 12$.

It is actually easy to enumerate all the possible ways:

$x = 1$ and $y = 11$
$x = 2$ and $y = 10$
$x = 3$ and $y = 9$
$x = 4$ and $y = 8$
$x = 5$ and $y = 7$
$x = 6$ and $y = 6$
$x = 7$ and $y = 5$
$x = 8$ and $y = 4$
$x = 9$ and $y = 3$
$x = 10$ and $y = 2$
$x = 11$ and $y = 1$