Sup web2.0, here's the problem:
At a local carnival, Mia is shooting at a row of 10 duck targets. Each time she shoots at a duck, the duck she shoots and any ducks immediately next to it fall down. In how many different ways can Mia shoot three ducks? (The order of the ducks also matters.)
idk I was doing something where I counted the total number ways to shoot 3 ducks and subtracting the arrangements where the two+ ducks shot were adjacent. That kind of ended up a mess, so anyone got some ideas?
Edit: To make it clear, I'm not looking for an answer, just a way of thinking about the problem if you could; ty :)
Mia can only shoot at Ducks 2-9, since there is no duck to the left of Duck 1 or to the right of Duck 10.
Therefore, the equation will be (9-2+1) ! which is 40320 ways.
Hope this helped!
I do not think that the rules are clear.
Order counts. If she knocks over duck number 2 then ducks one and duck 3 also fall. What order do they fall?
Also how many turns does Mia get?
Do misses pass as turns?
If she hits any duck from 2 to 9 then 3 ducks will go over. That is 8 possibilities.
If she hits duck one or duck 10 with the first shot then that is 2 down, the next shot that does not miss altogether will bring down at least 2 more, that is four or 5 in total.
So yea ... I do not know what is being asked.
@Melody, I think the problem is just saying that every time Mia shoots a duck, the ducks next to it fall down. Then, how many ways can Mia shoot three ducks? This should be clear now.