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

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,

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

step-by-step explanation:

always plug in x. in this case, x equals 4.

i'm going to first solve 4(2+3x)

1) 4(2+3(4)) use pemdas to solve.

i'm going to multiply whats in the parenthesis first, so 3 times 4=12.

2)4 (2+12) i'm then going to multiply 4 by both numbers in the parenthesis (because multiplying comes before adding)

4 times 2= 8.   4 times 12= 48

4) (8+48)= 56 (answer)

i'm then going to solve 8+12x and plug in 4 for x.

1) 8 +12(4)

multiplying comes before adding, so i'm gonna multiply 12 and 4.

12 times 4= 48.

2) 8+48=56

both equal the same!