- #1
Polymath89
- 27
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I have two basic questions about combinations. If there are e.g. 10 objects of which 3 are identical and you want to pick a group of 6 out of those 10, how many groups could you get in this case? I know how basic combinations work, but what if there are identical objects involved?
And my second question refers to these problems:
"In how many ways can you distribute 5 marbles in 3 identical baskets such that each basket has at least 1 marble?" and a variation of that with "3 distinct baskets".
Well for each question there are the possibilities that either one basket gets 3 and the other two 1 or two baskets get 2 and the remaining one 1. Now if the baskets are identical the solution for the 3-1-1 outcome is:
5C3*2C1*1C1/2!
and for the problem in which the baskets are distinct:
3C1*5C3*2C1*1C1
I understand the intuition behind the solution for the problem with the distinct baskets, but I don't understand why you have to divide by 2! on the problem with identical baskets, because I thought that the order does not matter, and 5C3*2C1*1C1 already reflects that the order does not matter, doesn't it?
I'd appreciate a short answer. Thanks in advance.
And my second question refers to these problems:
"In how many ways can you distribute 5 marbles in 3 identical baskets such that each basket has at least 1 marble?" and a variation of that with "3 distinct baskets".
Well for each question there are the possibilities that either one basket gets 3 and the other two 1 or two baskets get 2 and the remaining one 1. Now if the baskets are identical the solution for the 3-1-1 outcome is:
5C3*2C1*1C1/2!
and for the problem in which the baskets are distinct:
3C1*5C3*2C1*1C1
I understand the intuition behind the solution for the problem with the distinct baskets, but I don't understand why you have to divide by 2! on the problem with identical baskets, because I thought that the order does not matter, and 5C3*2C1*1C1 already reflects that the order does not matter, doesn't it?
I'd appreciate a short answer. Thanks in advance.