Three six-sided dice are rolled, one red, one green, one blue. How many ways are there to roll the three dice so that two of the dice have the same value and the other die has a value smaller than the common value on the other two?

Answer:

15

Step-by-step explanation:

Given that:

Three six-sided dices are to be rolled.

The number of ways to have the same value on the two dices is:

{(1,1) ,(2,2) ,(3,3) ,(4,4) , (5, 5 ) ,(6 ,6) }  = 6 ways

Thus; the objective is to determine the number of ways to a get smaller value on the third dice  while the two other dice have the same value.

To get started:

The number of ways for getting a value smaller than 1 = 0

The number of ways for getting a value smaller than 2 = 1

{i.e. (2,2,1)}

The number of ways for getting a value smaller than 3 = 2

{ i.e. (3,3,1) ,(3,3,2) }

The number of ways for getting a value smaller than 4 = 3

{ i.e. (4,4,1) , (4,4,2), (4,4,3)}

The number of ways for getting a value smaller than 5 = 4

{i.e. (5 ,5,1) ,(5,5,2) ,(5,5,3), (5,5,4)}

The number of ways for getting a value smaller than 6  = 5

{i.e. (6,6,1), (6,6,2), (6,6,3), (6,6,4), (6,6,5) }

Thus, the total number of ways = 0 + 1 + 2 + 3 + 4 + 5 = 15

Therefore, the total number of ways getting the value same on two dices and smaller value on the third dice = 15

Simplify x^-5

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