I need this
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.)
The sequence includes: 1,2,2,4,4,4.
We can do casework. If the seventh die roll is a 3,5,6, then the number of ways is the number of orderings.
\(\frac{7!}{2!3!}*3\) which simplifies to 1260.
because we need to divide by 3! for the 4s and 2! for the twos and multiply by 3 (using 3, 5, and 6).
Then, if the seventh roll is a 1, the sequence is an ordering of 1,1,2,2,4,4,4.
This, using the same reasoning as before, is
\(\frac{7!}{3!2!2!}\) which simplifies to 210.
If the seventh roll is a 2, then the sequence is 1,2,2,2,4,4,4, which is
\(\frac{7!}{3!3!}\) which simplifies to 140.
If the seventh roll is a 4, then the sequence is 1,2,2,4,4,4,4, which is
\(\frac{7!}{2!4!}\) which simplifies to 105.
Finally, we add all these together to get 1260+210+140+105=1715.
