+0

Counting

0
31
1

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.)

Feb 22, 2023

#1
+2602
+1

We need to split this problem up into 4 cases:

Case 1 - The last number is not a 1, 2, or 4: There are $${7! \over 3! 2!} = 420$$ cases

Case 2 - The last number is a 1: There are $${7! \over 3!2!2!} = 210$$ cases

Case 3 - The last number is a 2: There are $${7! \over 3!3!} = 140$$ cases

Case 4 - The last number is a 4: There are $${7! \over 4!2!} = 105$$ cases

So, there are $$420 + 210 + 140 + 105 = \color{brown}\boxed{875}$$ cases.

Feb 22, 2023