How Do You Solve Combination Problems with Duplicate Elements Efficiently?

[John112]

Are there efficient methods of findings the answer to combination problems like these?

The letters A, A, B and C are arranged in any order. How many DISTINCT sequences can we form?

If two letters weren't the same, then it would be simple. I can find the answer relatively easily since it's only 4 letters. Imagine if this was a sequence of 10 letters where two letters were the same. Then creating all possible combinations would be really inefficient and laboring. Is this any trick for problems where duplicate letters exist?

P.S. By the way the answer is 12. I found it by creating a tree diagram.

 

[LCKurtz]

Are there efficient methods of findings the answer to combination problems like these?

The letters A, A, B and C are arranged in any order. How many DISTINCT sequences can we form?

If two letters weren't the same, then it would be simple. I can find the answer relatively easily since it's only 4 letters. Imagine if this was a sequence of 10 letters where two letters were the same. Then creating all possible combinations would be really inefficient and laboring. Is this any trick for problems where duplicate letters exist?

P.S. By the way the answer is 12. I found it by creating a tree diagram.

If the A's were distinct, the number of arrangements would be 4! = 24. Now for each arrangement, wherever the two different A's are, there is another arrangement that is exactly the same except the A's are swapped. These two arrangements would be indistinguishable if the A's are the same, so that cuts the number of distinct arrangements in half, which gives 12.

In general if you have n objects of which k are the same, the number of arrangements is$$
\frac{n!}{k!}$$Do you see why?

 

[John112]

In general if you have n objects of which k are the same, the number of arrangements is$$
\frac{n!}{k!}$$Do you see why?

Yeah, I saw that. I wasn't sure whether that pattern of dividing by k factorial would hold for bigger sequences. I was trying to find a proof for that. Thanks for the helpful response. By the way is there already a proof for that general formula?

 

[LCKurtz]

Yeah, I saw that. I wasn't sure whether that pattern of dividing by k factorial would hold for bigger sequences. I was trying to find a proof for that. Thanks for the helpful response. By the way is there already a proof for that general formula?

Yes. Google permutations with some alike. One hit is:
http://dwb4.unl.edu/Chem/CHEM869N/CHEM869NMats/Permutations.html

The idea is that the k things that are alike would have given k! permutations but only one if they were the same.

 

[CellCoree]

I can say that there are indeed efficient methods for finding the answer to combination problems like the one presented. One such method is known as the "combination formula," which is used to calculate the number of distinct combinations that can be formed from a set of objects, taking into account repeating elements. In this particular case, the formula would be n! / (n1! * n2! * ... * nk!), where n is the total number of objects, and n1, n2, etc. represent the number of repeated elements.

In the example given, n = 4 (A, A, B, C), so the formula would be 4! / (2! * 1! * 1!) = 12, which matches the answer found by creating a tree diagram. This formula is much more efficient than manually creating all possible combinations, especially for larger sets of objects.

Another efficient technique for solving combination problems is the use of generating functions, which represent the different combinations as coefficients in a polynomial equation. This method is particularly useful for more complex problems with multiple sets of objects and repeating elements.

In conclusion, as a scientist, I can assure you that there are efficient methods for solving combination problems, even when duplicate elements are present. These techniques not only save time and effort but also showcase the power of mathematical concepts in problem-solving.