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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.)

 Jun 21, 2023
 #1
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The answer is 66.

 Jun 21, 2023
 #2
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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.)  

 

There are two ways to obtain a product of 2500 using factors between 1 and 6 inclusive.  

 

One way is             5 • 5 • 5 • 5 • 2 • 2  

The other way is    5 • 5 • 5 • 5 • 4 • 1  

 

I admit I had to look that up.  

I searched on google for prime factorization of 2500.  

 

I don't know how to use permutations to mix those other numbers  

in there with all those fives, so I would have to use brute force.  

 

225555    522555    552255    555225    555522  

252555    525255    552525    555252   

255255    525525    552552   

255525    525552   

255552  

 

If I didn't miss any, there are 15 with 5's and 2's.  

 

There would be another 30 with 5's, a 4, and a 1. 

This is because there would be 15 with the 1 first, 

then another 15 with the 4 first.  I think that's right.  

I'm not going to write them all down.  If you want to  

write them all down, that would be okay with me. 

I think your teacher would be impressed. ;-)  

 

Anyway, the total number of sequences,  

according to my feeble calculations, seems to be 45 sequences.  

.

 Jun 22, 2023
 #3
avatar+2437 
-7

Hi Ron,

 

\(\small \text {Here's the formula to solve for unique distinguishable permutation counts where the set of elements has one or more subsets of identical elements }\\ \small \text {(Each redundancy is factored out.) }\\ \; \dfrac {^n \mathrm P_r} {\mathrm d_1! \cdot \mathrm d_2! \; \cdots \mathrm d_x! } | \text { Where } (\mathrm d_x) \text{ is the number of duplicates for each subset of duplicates.}\\ \small \text {Because this solution does not require subset permutation counts, } (^n \mathrm P_r) \text { simplifies to } \mathrm (\mathrm P!) \\ \dfrac {\mathrm P!} {\mathrm d_1! \cdot \mathrm d_2! \; \cdots \mathrm d_x! }\Rightarrow \dfrac{6!}{4! * 2!} = 15 \; \; \text{ and } \; \dfrac{6!}{4! } = 30 \)

 

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It’s wonderful that you’ve become a member, Ron. I’m sure your cat is pleased with your user name. Meow

 

 

GA

--. .- 

GingerAle  Jun 22, 2023
 #4
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0

Hey Ginger!  

 

It's so great to hear from you.  I haven't been seeing postings of yours for quite a while, and I was afraid I never would again.  Thank you for that explanation about permutations.  I agree that Bosco the Chocolate Bear would have been pleased with my username.  Or is pleased, if he's watching from cat heaven. 

 

This is something funny.  It turns out that my username here is a cat I used to have, and my user name on eBay is a cat my grandma used to have.  I didn't plan it that way, and the coincidence just dawned on me a couple of days ago. 

 

Ron  

Bosco  Jun 23, 2023
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