Just to start off, I want you to read my answer and explanation, and to make sure you do that, I will be solving a SIMILAR question, but not exactly. Here is the question I will be solving:
The following cards are dealt to FOUR people at random, so that everyone gets the same number of cards. What is the probability that everyone gets a red card?
[There are FOUR red cards and FOUR blue cards]
To start off, if we are going to calculate probability, that means that there is going to be a fraction in the form of:
Let's find the total first. To start, we have 8 cards for the first person, 7 cards for the second, 6 cards for the third and 5 for the fourth, multiplying gives us:
8*7*6*5 total possibilities (we aren't going to calculate this just yet)
Then, we need to find the probability that everyone gets a red card. There are only 4 cards, and we can put them respectively to each of them 4! different ways (4 * 3 * 2 * 1). Therefore the probability is:
This simplifies to:
This simplifies to:
Note: This is not the answer to your question, but the answer to a similar question. I advise you to read this explanation, and if you find any errors, please tell me (as counting isn't my strong subject, there might be something wrong).
Alright, so here's the deal, I think I might be able to do this problem, but no guarantee that it is going to be right. I advise you to check my work afterward, to make sure that it is right:
So, we first know that there are 10 dogs, and we have to split them up into a 3 group, a 5 group, and a 2 group. However, one dog is already set in stone for the 2 dog group, as well as another dog set in stone for the 5 dog group.
We first pick the 3 dog group. We can do this by:
C(8,3) which means (choose 3 of 8). C(8,3) is 56.
Then, for the 5 dog group, we so far have 1 dog already established, then from the remaining 5 dogs left, we can pick 4 of them.
C(5,4) = 5 ways here.
The last group is already decided because we already picked the other two.
When we do 56 * 5 = 280, so I believe there are 280 ways to do this. (However, please REMEMBER, I might be incorrect [ and probably am] )