Question

A person rolls a standard six-sided die 9 times. In how many ways can he get 3 fours, 5 sixes, and 1 two?

Answer 1

Answer:

There are 5040 ways he get 3 fours, 5 sixes, and 1 two.

Step-by-step explanation:

Given : A person rolls a standard six-sided die 9 times.

To find : In how many ways can he get 3 fours, 5 sixes, and 1 two?

Solution :

A person rolls a standard six-sided die 9 times.

So, total number of ways die roll is 9! ways.

In die 4 comes 3 times.

So, ways of getting 4 is 3!

In die 6 comes 5 times.

So, ways of getting 6 is 5!

In die 2 comes 1 times.

So, ways of getting 2 is 1!

Total number of ways is given by,

$T=\frac{10!}{3!\times 5!\times 1!}$

$T=\frac{10\times 9\times 8\times 7\times 6\times 5!}{3\times 2\times 5!\times 1}$

$T=10\times 9\times 8\times 7\times$

$T=5040$

Therefore, there are 5040 ways he get 3 fours, 5 sixes, and 1 two.