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# Counting

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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

#1
+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
+1252
+1

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