+216 A standard six-sided die is rolled $7$ times. You are told that among the rolls, there was one $1,$ one $2$, one $3$, one $4$, one $5$, and two $6$s. How many possible sequences of rolls could there have been? (For example, 2, 3, 4, 6, 6, 1, 5 is one possible sequence.)
+1953 We can do this using a tactic.
First, let's find the number of ways we can arrange the 7 rolled numbers.
We just have 7!=7∗6∗5∗4∗3∗2∗1
However, we can't just end here. Since there are two 6s, we must account for the duplicates.
We do this by dividing the entire thing by 2!
Thus, we just have
7!/2!=2520
So our answer is just 2520.
Thanks! :)
+1953 We can do this using a tactic.
First, let's find the number of ways we can arrange the 7 rolled numbers.
We just have 7!=7∗6∗5∗4∗3∗2∗1
However, we can't just end here. Since there are two 6s, we must account for the duplicates.
We do this by dividing the entire thing by 2!
Thus, we just have
7!/2!=2520
So our answer is just 2520.
Thanks! :)
