Formula for combinations with repeting objects

In summary, the conversation discusses the picking of two letters from a set of four (A, B, C, D) for which there are six possible combinations. However, if one of the letters is replaced by a new letter (A2), there are only four unique combinations due to the indistinguishability of two combinations (A1A2 and A2A1). The formula for combinations in this case is not mentioned, but it is noted that for permutations, the formula would be 6/2=3.
  • #1
musicgold
304
19
Hi,

I am not sure which formula will allow me to get the following results.

If I have to pick two letters from A, B, C, D, then there are 6 possible combinations:
AB
AC
AD
BC
BD
CD.

Now assume that letter D has turned into another A, i.e. A2. Now my combinations have changed to
A1B
A1C
A1A2
BC
BA2
CA2

So I eliminate BA2, and CA2 and I have 4 unique combinations.

I know that in the case of permutations with repetitions, we divide the nPr formula by the factorials of the frequency of the letters, but I am not sure how it is done with combinations. What formula can use for combinations?

Thanks.
 
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  • #2
musicgold said:
Hi,

I am not sure which formula will allow me to get the following results.

If I have to pick two letters from A, B, C, D, then there are 6 possible combinations:
AB
AC
AD
BC
BD
CD.
Yes, that is
[tex]\left(\begin{array}{c} 4 \\ 2\end{array}\right)= \frac{4!}{2! 2!}= 6[/tex]

Now assume that letter D has turned into another A, i.e. A2. Now my combinations have changed to
A1B
A1C
A1A2
BC
BA2
CA2

So I eliminate BA2, and CA2 and I have 4 unique combinations. [/quote]
Okay, with the "1" appended you still have 4 objects but without that, you cannot distinguish between combinations where the "A"s have been swapped. However, here "A1A2" and "A2A1" which are two distinct permutations are NOT two distinct combinations.
If you were counting permutations, then it would be 6/2= 3 but with that "A1A2" and "A2A1" put back in there are 4 combinations.

I know that in the case of permutations with repetitions, we divide the nPr formula by the factorials of the frequency of the letters, but I am not sure how it is done with combinations. What formula can use for combinations?

Thanks.
 
  • #3
Thanks HallsofIvy.


Okay, with the "1" appended you still have 4 objects but without that, you cannot distinguish between combinations where the "A"s have been swapped. However, here "A1A2" and "A2A1" which are two distinct permutations are NOT two distinct combinations.
If you were counting permutations, then it would be 6/2= 3 but with that "A1A2" and "A2A1" put back in there are 4 combinations.

Your answer is not clear to me. As I said, I know the formula for the combination 4C2.
I am interested in the formula for the case where two letters are the same.
 

1. What is the Formula for Combinations with Repeating Objects?

The formula for combinations with repeating objects is n^r, where n is the number of objects and r is the number of times each object can be repeated. This formula is used to calculate the number of unique combinations that can be created from a set of objects with repetition allowed.

2. When is the Formula for Combinations with Repeating Objects used?

This formula is used in situations where objects can be repeated in a combination, such as when selecting a certain number of objects from a larger set with replacement. It is commonly used in probability and statistics to calculate the number of possible outcomes in a given scenario.

3. Can the Formula for Combinations with Repeating Objects be applied to real-life problems?

Yes, this formula can be applied to real-life problems such as choosing a combination of toppings for a pizza or selecting a combination of cards from a deck. It can also be used in business and marketing scenarios, such as determining the number of possible product combinations for a new line of clothing.

4. How is the Formula for Combinations with Repeating Objects different from the Formula for Combinations without Repetition?

The formula for combinations without repetition is n!/(r!(n-r)!), where n is the number of objects and r is the number of objects in each combination. This formula is used when objects cannot be repeated in a combination. The formula for combinations with repetition, n^r, takes into account the possibility of objects being repeated in a combination.

5. How can the Formula for Combinations with Repeating Objects be calculated?

The formula for combinations with repeating objects can be calculated using a calculator or by hand. Simply raise the number of objects (n) to the power of the number of repetitions (r). For example, if there are 4 objects and each one can be repeated 3 times, the calculation would be 4^3 = 64 possible combinations.

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