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A standard die with six sides is rolled 5 times. I get one 1, one 4 and three 3's. How many possible sequences of rolls are there? (For example, 3, 3, 1, 3, 4 is one possible sequence.)

 

I just did \(6^5\) but that's incorrect, maybe I just don't understand the problem because I don't see any restrictions :P.

 Dec 23, 2019
 #1
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There are C(5,3) = 10 ways to choose the position of the 3s, then C(5,1) = 5 ways to choose the position of the 4, and C(5,1) = 5 ways to choose the position of the 1, so the number of sequences is \(10 \cdot 5 \cdot 5 = \boxed{250}\)

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 Dec 23, 2019
 #2
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If we think of the numbers as letters, there are 5! ways to arrange them but since 3 is repeated 3 times, we divide by 3! So 5!/3! = 20.

 Dec 23, 2019
 #3
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{1, 3, 3, 3, 4} | {1, 3, 3, 4, 3} | {1, 3, 4, 3, 3} | {1, 4, 3, 3, 3} | {3, 1, 3, 3, 4} | {3, 1, 3, 4, 3} | {3, 1, 4, 3, 3} | {3, 3, 1, 3, 4} | {3, 3, 1, 4, 3} | {3, 3, 3, 1, 4} | {3, 3, 3, 4, 1} | {3, 3, 4, 1, 3} | {3, 3, 4, 3, 1} | {3, 4, 1, 3, 3} | {3, 4, 3, 1, 3} | {3, 4, 3, 3, 1} | {4, 1, 3, 3, 3} | {4, 3, 1, 3, 3} | {4, 3, 3, 1, 3} | {4, 3, 3, 3, 1} (total: 20)

 Dec 23, 2019

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