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A standard six-sided die is rolled $7$ times.  You are told that among the rolls, there was one $1,$ one $2$, one $3$, one $4$, one $5$, and two $6$s.  How many possible sequences of rolls could there have been?  (For example, 2, 3, 4, 6, 6, 1, 5 is one possible sequence.)

 Jun 20, 2024

Best Answer 

 #1
avatar+1252 
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We can do this using a tactic. 

First, let's find the number of ways we can arrange the 7 rolled numbers. 

We just have \(7! = 7*6*5*4*3*2*1\)

 

However, we can't just end here. Since there are two 6s, we must account for the duplicates.

We do this by dividing the entire thing by \(2!\)

 

Thus, we just have

\(7! / 2! = 2520 \)

 

So our answer is just 2520. 

 

Thanks! :)

 Jun 20, 2024
 #1
avatar+1252 
+1
Best Answer

We can do this using a tactic. 

First, let's find the number of ways we can arrange the 7 rolled numbers. 

We just have \(7! = 7*6*5*4*3*2*1\)

 

However, we can't just end here. Since there are two 6s, we must account for the duplicates.

We do this by dividing the entire thing by \(2!\)

 

Thus, we just have

\(7! / 2! = 2520 \)

 

So our answer is just 2520. 

 

Thanks! :)

NotThatSmart Jun 20, 2024

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