Respuesta :

Answer:

20

Step-by-step explanation:

Data provided in the question:

Number of times a die is rolled = 5

Number repeating

4 - 1 times

5 - 1 times

6 - 3 times

Now,

The number of possible sequences of rolls can be the factorial of number of times the dice is rolled to the factorial of the number to times a number is repeated

Therefore,

The number of possible sequences of rolls = [tex]\frac{5!}{1!\times1!\times3!}[/tex]

or

The number of possible sequences of rolls = [tex]\frac{5\times4\times3!}{1\times1\times3!}[/tex]

or

The number of possible sequences of rolls = 20

s. How many possible sequences of rolls could there have been? (For example, $6,5,6,6,4$ is one possible sequence.)

Respuesta :

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s.

We need to find possible sequences of rolls could there have been.

Total numbers to arrange (i.e.,4,5,6,6,6)= 5

Repleted numbers (i.e., 6,6,6) = 3

[tex]5P_5=\dfrac{5!}{(5-5)!}=5\times 4\times 3\times 2\times 1=120[/tex]

[tex]3P_3=\dfrac{3!}{(3-3)!}=3\times 2\times 1=6[/tex]

Total possible ways are

[tex]Total=\dfrac{5P_5}{3P_3}[/tex]

[tex]Total=\dfrac{120}{6}[/tex]

[tex]Total=20[/tex]

Therefore, the total possible sequences of rolls is 20.

s. How many possible sequences of rolls could there have been? (For example, $6,5,6,6,4$ is one possible sequence.)

Respuesta :

Answer:

20

Step-by-step explanation:

Data provided in the question:

Number of times a die is rolled = 5

Number repeating

4 - 1 times

5 - 1 times

6 - 3 times

Now,

The number of possible sequences of rolls can be the factorial of number of times the dice is rolled to the factorial of the number to times a number is repeated

Therefore,

The number of possible sequences of rolls = [tex]\frac{5!}{1!\times1!\times3!}[/tex]

or

The number of possible sequences of rolls = [tex]\frac{5\times4\times3!}{1\times1\times3!}[/tex]

or

The number of possible sequences of rolls = 20

s.

We need to find possible sequences of rolls could there have been.

Total numbers to arrange (i.e.,4,5,6,6,6)= 5

Repleted numbers (i.e., 6,6,6) = 3

[tex]5P_5=\dfrac{5!}{(5-5)!}=5\times 4\times 3\times 2\times 1=120[/tex]

[tex]3P_3=\dfrac{3!}{(3-3)!}=3\times 2\times 1=6[/tex]

Total possible ways are

[tex]Total=\dfrac{5P_5}{3P_3}[/tex]

[tex]Total=\dfrac{120}{6}[/tex]

[tex]Total=20[/tex]

Therefore, the total possible sequences of rolls is 20.