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# need help

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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