Possible webpage title: Understanding Combinations with Identical Objects

In summary, the conversation discusses two basic questions about combinations involving identical objects. The first question is about picking a group of objects from a larger group, while the second question involves distributing a certain number of objects into baskets. The solutions for these problems involve using combinations and considering the possibility of at least one basket not having a certain number of objects. For the problem with identical baskets, there is an additional step of dividing by 2! to account for the order not mattering.
  • #1
Polymath89
27
0
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.
 
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  • #2
Hey Polymath89.

The easiest way to think about this is to consider the complement where at least one basket doesn't have one marble.

This is given by either (5,0,0) (4,1,0) or (3,2,0) combinations. The first can happen 3C1 times and the other two can happen 3C2 times.

In total we get 3C1 + 2*3C2 = 3 + 12 = 15 different ways of not getting at least 1 in every slot.

I'll think about the number of possibilities in a sec.
 

Related to Possible webpage title: Understanding Combinations with Identical Objects

1. What are combinations?

Combinations refer to the different ways in which a set of objects can be selected or arranged without taking the order into consideration.

2. How do you calculate the number of combinations?

The number of combinations can be calculated using the formula nCr = n! / r!(n-r)!, where n is the total number of objects and r is the number of objects selected.

3. What is the difference between combinations and permutations?

Combinations do not take the order of selection into account, while permutations do. In other words, combinations focus on the group of objects being selected, while permutations focus on the arrangement of those objects.

4. Can combinations be repeated?

No, combinations cannot be repeated. In other words, once an object is selected, it cannot be selected again in the same combination.

5. How are combinations used in real life?

Combinations are used in many fields, such as mathematics, statistics, and computer science. They are used to solve problems involving selecting a specific group of items from a larger set, such as in genetics, probability, and data analysis.

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