Probability proof by combinatorial argument

There are (^m_x) ways of doing this. However, x can be any number from 0 to r, so you have to sum over all possible values of x. This gives you the desired result.In summary, by using a combinatorial argument, we can prove that for r \leq n and r \leq m, the number of ways to pick r items from a set of n + m items is equal to the sum of the permutations of picking 0, 1, 2, ..., r items from two subsets of m and n items respectively. This is because each choice on the right side corresponds to one of the choices on the left side, leading to the desired result.
  • #1
Proggy99
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0

Homework Statement


By a combinatorial argument, prove that for r [tex]\leq[/tex] n and r [tex]\leq[/tex] m,
[tex](^{n+m}_{r})[/tex] = [tex](^{m}_{0})(^{n}_{r})[/tex] + [tex](^{m}_{1})(^{n}_{r-1})[/tex] + ... + [tex](^{m}_{r})(^{n}_{0})[/tex]


Homework Equations





The Attempt at a Solution


I need some direction on how to start this problem. It is the only homework problem I am not sure of how to approach it.
 
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  • #2
okay, so here is my attempt at beginning the solution

The left side gives the number of ways that you can pick r total items from a set made up of two subsets of items with m and n items in each subset.

The right side gives a series of permutations including how to pick no items from the first subset and all r items from the second, then how to pick 1 item from the first subset and the rest from the second, then 2 items from the first subset and the rest from the second, and continuing on until you are choosing all r items from the first subset and no items from the second.

I am just not sure how to go about putting this in proof form
 
  • #3
So a choice on the right side must correspond to one of the choices on the left side and vice versa, right? I think that's exactly what you want to say, and I would call it a 'proof'.
 
  • #4
You can argue that in order to pick r items from m + n items, you have to pick x from the m items and r - x from the n items.
 

1. What is a combinatorial argument in probability proof?

A combinatorial argument in probability proof involves using counting techniques to determine the number of possible outcomes in a given situation. It is a way to mathematically prove the probability of an event occurring.

2. How is a combinatorial argument used in probability proof?

A combinatorial argument is used in probability proof by considering all the possible ways an event can occur and determining the probability of each outcome. These probabilities are then added together to get the overall probability of the event occurring.

3. What are some common counting techniques used in combinatorial arguments?

Some common counting techniques used in combinatorial arguments include the multiplication principle, combinations, permutations, and binomial coefficients.

4. Can combinatorial arguments be used for both discrete and continuous probability distributions?

Yes, combinatorial arguments can be used for both discrete and continuous probability distributions. However, they are more commonly used for discrete distributions where the outcomes are countable.

5. What are the advantages of using a combinatorial argument in probability proof?

Using a combinatorial argument in probability proof allows for a clear and systematic approach to determining the probability of an event. It also provides a way to verify the calculated probability and can be applied to a wide range of probability problems.

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