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Catherine rolls a standard 6-sided die six times. If the product of her rolls is \(2500\) then how many different sequences of rolls could there have been? (The order of the rolls matters.)

 Jul 6, 2023
 #1
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150        

 Jul 6, 2023
 #2
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Explanation: The prime factorization of \(2500\) is \(2^2*5^4\), so Catherine will roll 2 twos and 4 fives. Now we just need to deal with the amount of ways to reorganize those rolls. If we have 6 slots:

 

[] [] [] [] [] [], and place the 2 twos in them, the other 4 will be automatically placed.

 

And there are \(\frac{6*5}{2}\) 15 ways of doing this (you can also use the combination formula for this). That gives us \(\boxed{15}\) ways of doing so.

 Jul 6, 2023
 #5
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>>>>>  so Catherine will roll 2 twos and 4 fives  <<<<<   

 

            instead of 2 twos, there could be 1 one and 1 four.  

.

Bosco  Jul 6, 2023
 #3
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2500 = 2^2 * 5^4

 

There are only 2 combinations as follows:

 

145555 ==6! /4! ==30 permutations

225555==6!/4!2!= 15 permutations

 

Total ==30 + 15 ==45 different sequences of rolls.

 Jul 6, 2023
 #4
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Oh you're right @guest, completely forgot about that part lol. Good job!

SoulKingBrook  Jul 6, 2023

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