How many ways to roll a die seven times and obtain a sequence of outcomes with three 1s, two 5s, and two 6s?
Ans: When I was thinking of a way to decompose the problem I first thought that if a die is getting rolled seven times then each roll is independent of the other thus: $6^7$ for all possible outcomes… then I was having trouble with how to take care of the constraint. The solution given is: $$\frac{7!}{3!2!2!}$$
The section I am covering right now is about repetition, but I do not see at al how this could be the result. I thought I am supposed to find the outcomes WITH three 1s, two 5s, and two 6s? Isn't dividing those sequemces out doing the exact opposite and now we are finding all the arrangements WITHOUT three 1s, two 5s, and two 6s?
Best Answer
Another way to think about it goes like this: So you roll seven times. First, choose 3 spots for the 1s,$$\binom{7}{3}.$$ Next, choose 2 spots for the 5s. We already used up 3, so this is $$\binom{4}{2}.$$ Finally, choose the last to spots for the 6s, $$\binom{2}{2}.$$ This gives $$\binom{7}{3}\binom{4}{2}\binom{2}{2} = \frac{7!\,4!\,2!}{3!\,4!\,2!\,2!\,2!\,0!} = \binom{7}{3,2,2}.$$