# In how many ways can one get a sum greater than 8, if two cubes, having numbers 1 to 6 are rolled?

Mar 17, 2017

There are ten number of ways in which one could get a sum that is greater than $8$.

#### Explanation:

As each cube contains number $1$ to $6$ (i.e. $6$ options), there are $6 \times 6 = 36$ options.

These options start from $\left(1 , 1\right)$ and go on till $\left(6 , 6\right)$.

(1,1),(1,2),....(1,5),*1,6),(2,1),....(2,6),(3,1),....(6,6)

Observe that we can get a sum maximum of $12$

What is desired is the sum that is greater than $8$, i.e. $9$, $10$, $11$ and $12$.

You can get $12$ in only one way i.e. $\left(6 , 6\right)$.

$11$ can be got in two ways $\left(5 , 6\right)$ and $\left(6 , 5\right)$

$10$ can be got in three ways $\left(4 , 6\right)$, $\left(5 , 5\right)$ and $\left(6 , 4\right)$ and

$9$ can be got in four ways $\left(3 , 6\right)$, $\left(4 , 5\right)$, $\left(5 , 4\right)$ and $\left(6 , 3\right)$

Hence, there are ten number of ways in which one could get a sum that is greater than $8$.