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

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Answer 1 · 2017-03-17T04:37:31

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

As each cube contains number $1$ to $6$ (i.e. $6$ options), there are $6xx6=36$ options.

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

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

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. $(6,6)$ .

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

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

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

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

In how many ways can one get a sum greater than 88, if two cubes, having numbers 11 to 66 are rolled?