+0  
 
+1
106
2
avatar

Plz help

 

A standard six-sided die is rolled 7 times.  You are told that among the rolls, there was one 1, two 2's, and three 4's.  How many possible sequences of rolls could there have been?  (For example, 5, 1, 2, 2, 4, 4, 4 is one possible sequence.  4, 1, 4, 4, 2, 1, 1 is one possible sequence.)

 Apr 8, 2023
 #1
avatar+81 
+1

So I'm not the best at math, but I'll give this a shot. We have the set {1,2,2,4,4,4}. First off, we have 6! = 720, not counting any duplicates. Because there are 3 copies for each time we rearrange the set, (because some of them are us just rearranging the fours), we divide 720 by 3, to get 240. We repeat this process with the two 2s, and get 120. Now, I'm not 100% sure about this answer, so Melody (or anybody else), if you see this and it's wrong, could you please correct me? Thank you!

 Apr 8, 2023
edited by moneydude242  Apr 8, 2023
edited by moneydude242  Apr 8, 2023
 #2
avatar
+1

1122444 ( 210 ) , 1222444 ( 140 ) , 1223444 ( 420 ) , 1224444 ( 105 ) , 1224445 ( 420 ) , 1224446 ( 420 ) , >>Total combinations = 6

 

>>Total permutations(totals in brackets) = 1,715 permutations - number of possible sequences of rolls.

 Apr 8, 2023

2 Online Users