When selecting $r$ objects from $n$ distinct objects with repetition allowed and ordering does not matter, the problem is equivalent to finding the number of combinations with repetition. This can be solved using the "stars and bars" theorem.

Formula:

The number of ways to choose $r$ objects from $n$ distinct objects with repetition allowed is given by the combination formula:

$\text{Number of selections} = \binom{n + r - 1}{r}$

Explanation:

• $n + r - 1$ represents the total number of items (including the repetitions and the original objects).
• $r$ is the number of objects being chosen.
• The binomial coefficient $\binom{n + r - 1}{r}$ is calculated as:

$\binom{n + r - 1}{r} = \frac{(n + r - 1)!}{r! \cdot (n - 1)!}$

This formula counts the number of ways to place $r$ identical items into $n$ distinct categories, where some categories may be empty (which corresponds to repetition being allowed).

Example:

If you have 3 distinct objects and you want to select 2 objects with repetition allowed, the number of ways would be:

$\binom{3 + 2 - 1}{2} = \binom{4}{2} = \frac{4!}{2! \cdot 2!} = \frac{24}{4} = 6$

There are 6 ways to choose 2 objects with repetition allowed from 3 distinct objects.

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Do you want more details or have any questions?

Here are 5 related questions:

1. How would the formula change if repetition is not allowed?
2. How does the concept of "stars and bars" work in combinatorics?
3. What is the difference between permutations with repetition and combinations with repetition?
4. Can this formula be used in problems involving partitioning objects into groups?
5. How would you calculate the number of ways to distribute $r$ identical objects into $n$ distinct boxes?

Tip: Remember that when dealing with combinations where repetition is allowed, think of it as placing indistinguishable stars (chosen objects) into distinguishable boxes (distinct objects).