As I study for the P exam, my mind always gets distracted by things that my book either doesn't address or won't address until later (which I don't have the patience to find out, because it will bug me so much!) Here are two questions that came up.

1) We know that when counting Permutations, and some objects are identical in the set, we simply do n!/(r1! * r2! ...) where n is the number of objects, and r1 objects are identical to each other, r2 items are identical to each other, and so on.

Using Combinatorics, we know that (n choose r) gives us the number of possible combinations of n objects taken r at a time (when the order of selection is irrelevant).

Now, what if some of the objects from the original group of n where in fact identical? For example, We have want to work with the word PEPPER, and we want to determine the number of possible combinations when we take 4 items at a time. It seems a little different. My intuition tells me that we just divide by 3! and 2! ...? But my intuition is normally wrong. Can someone generalize this using n's and r's? :wacko:

2) We know that the notation (n choose n1, n2, ..., nr) represents the number of possible divisions of n distinct objects into rdistinctgroups, of respective sizes n1, n2... and the number of items in all of the groups sum to n. So what determines if one group is "indistinguishable" from another? For example, 10 children are going to be divided into an A and B team. How many different divisions are there? It's simply (10 choose 5, 5) = 10!/(5! * 5!). But if we just say, 10 children are going to divide themselves into two teams of 5. Now the problem is different, and the answer is [10!/(5! * 5!)]/2!

Why is the second one different? Is it because we labeled the groups in the first example? What if we didn't label the groups, and just said that they were going to divide themselves into a group of 6 and a group of 4. Is the fact that these two groups are different in number by definition make them different? I.e., the "6" group and the "4" group. If that's the case, then the only time the second example's thought process comes into play is when 1) we don't "label" the groups beforehand and 2) ALL the groups are the exact same size, i.e., n1 = n2 = n3 = ... = nr? and 2) :shocked: And if THAT'S the case, I'm sure there's alot of linear algebra that could be done to figure out just how many times this is possible.. for example, from 10 items, you can do this with groups of 2 and 5, but not 3's because there is no linear combination of 3's using integers that can sum to 10? But don't ask me, I barely passed linear algebra.

Thanks for the input on all or any parts of this post... It's been bothering me for a while!