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Catherine rolls a $6$-sided die five times, and the product of her rolls is $1200.$ How many different sequences of rolls could there have been? (The order of the rolls matters.)

 Jul 8, 2020
 #1
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1200  factors  as      2^4 * 3  * 5^2 =   2 * 2 * 2 * 2 * 3 * 5 * 5  =  4 * 4 * 3 * 5 * 5

 

So....the number of different identifiable roll sequenes  =

 

        5!                        120      

    _______  =            ____    =     30

     2!  * 2!                      4

 Jul 9, 2020
 #2
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mmm that answers wrong

Guest Jul 9, 2020
 #3
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If I understand your question, these are all the sequences possible on a roll of 1 die 5 times:

 

24556 , 24565 , 24655 , 25456 , 25465 , 25546 , 25564 , 25645 , 25654 , 26455 , 26545 , 26554 , 34455 , 34545 , 34554 , 35445 , 35454 , 35544 , 42556 , 42565 , 42655 , 43455 , 43545 , 43554 , 44355 , 44535 , 44553 , 45256 , 45265 , 45345 , 45354 , 45435 , 45453 , 45526 , 45534 , 45543 , 45562 , 45625 , 45652 , 46255 , 46525 , 46552 , 52456 , 52465 , 52546 , 52564 , 52645 , 52654 , 53445 , 53454 , 53544 , 54256 , 54265 , 54345 , 54354 , 54435 , 54453 , 54526 , 54534 , 54543 , 54562 , 54625 , 54652 , 55246 , 55264 , 55344 , 55426 , 55434 , 55443 , 55462 , 55624 , 55642 , 56245 , 56254 , 56425 , 56452 , 56524 , 56542 , 62455 , 62545 , 62554 , 64255 , 64525 , 64552 , 65245 , 65254 , 65425 , 65452 , 65524 , 65542 , Total =  90

 Jul 9, 2020

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