Help!! Assume that there are 5 dogs in a box: 3 with legs and 2 without legs. A dog is removed from the box and it is noted whether or not it has legs. It is then placed in a different cage. This occurs 2 more times. How many possible outcomes are there?

Question

anonymous · Answer

I'm guessing permutations, but I've completely forgotten how to do them!

zzr0ck3r · Answer

do we take in to acount that the dog is both alive and dead?

anonymous · Answer

You would then have to apply schrodinger's equation which is a differential equation. That would make things much more difficult!

anonymous · Answer

you are choosing 3 out of a set of 5
number of ways to do this is called "five choose 3" written as $\dbinom{5}{3}$ and computed via 
$$\dbinom{5}{3}=\frac{5\times 4}{2}=10$$

anonymous · Answer

satellite73 I don't think the dogs are meant to be individually identifiable, i think the suggestion is that they only differ in regards to whether or not they have legs.

anonymous · Answer

I'm not exactly sure what constitutes an outcome. If we are just talking about what get's 'noted' down then theres only 2*2*2-1 =7. The first dog can be of 2 types: legs or no legs. The same is true of the 2nd and 3rd dogs, well almost, you have to minus 1 off because all three observed dogs can't be legless as there are only two legless dogs in the box to start with.

anonymous · Answer

So, it would be 7*3 being 21?