I am looking to verify the following information from a set of knitting needles for fun. The set contains 8 pairs of needles (they have different sizes), 3 cords and 1 connector. They claim that this results in more than 500 combinations.
For knitting you need two needles and at least one cord. You can connect any two cords with the connector.
I tried to simplify the problem by first getting the number of cord combinations cc. I think it is cc = 3 + 3 = 6 because if we call the cords A B and C, we can use them on their own (3) or connect: A-B, B-C, A-C.
Usually, you would use two needles of the same size. I think under that assumption the number of combinations is: cc * (number of same pairs) = 6 * 8 = 48. Is that right?
But I guess they also include using different size of needles. So, I would have to find the probability of picking two needles from the set of 16 needles. However, since there are duplicates I am not super sure what the right solution is.
What I tried:
I found a formula on Wikipedia to get the multiset number, where the number of multisets of cardinality k, with elements taken from a finite set of cardinality n. My problem seems to fit in this. I have the set {1,2,3,4,5,6,7,8} (simplifying the needles sizes to integers). And I want to take sets of cardinality 2 from this. So: (9 choose 2)* cc = 216.
Also if I imagine picking the first needle I have 8 possibilities. If I want to pick the same size for the second 8cc=48 otherwise 87/2*cc=168. Together 216.
What am I doing wrong? Or is it the company's advertisement that has a mistake?