Suppose a game has [imath]c[/imath] items and [imath]s[/imath] attachments for those items. The player can attach [imath]0[/imath] or [imath]1[/imath] of the attachments to each item. How many configurations are there?
I first tried to model the problem using the twelvefold way. I believe the interpretation would be in the direction of counting functions from the set of attachments to the set of items. However, I reasoned that none of these formulas can be correct, because neither all items nor all attachments must be used, and so missing both left totality and right totality, I should not be using functions in either direction.
I then looked up this old question extending the twelvefold way to relations. It seem to me that I want the first table in the answer, because each item is uniquely identifiable and each attachment is uniquely identifiable. As mentioned, totality in both directions is out the window, but there does seem to be both left uniqueness and right uniqueness, since each item can have at most [imath]1[/imath] attachment by stipulation and each attachment can be present on at most [imath]1[/imath] item by the nature of the case. So I believe the formula I need is...
[math]\sum_{k = 0}^{\min{\{c,\ s}\}} {c \choose k}{s \choose k}k![/math]
It is reassuring that this formula treats [imath]c[/imath] and [imath]s[/imath] symmetrically, since I assigned the same properties to both at each stage. Is this the correct formula for the problem? If not, then where have I gone wrong modeling this problem in terms of relation counting?
I first tried to model the problem using the twelvefold way. I believe the interpretation would be in the direction of counting functions from the set of attachments to the set of items. However, I reasoned that none of these formulas can be correct, because neither all items nor all attachments must be used, and so missing both left totality and right totality, I should not be using functions in either direction.
I then looked up this old question extending the twelvefold way to relations. It seem to me that I want the first table in the answer, because each item is uniquely identifiable and each attachment is uniquely identifiable. As mentioned, totality in both directions is out the window, but there does seem to be both left uniqueness and right uniqueness, since each item can have at most [imath]1[/imath] attachment by stipulation and each attachment can be present on at most [imath]1[/imath] item by the nature of the case. So I believe the formula I need is...
[math]\sum_{k = 0}^{\min{\{c,\ s}\}} {c \choose k}{s \choose k}k![/math]
It is reassuring that this formula treats [imath]c[/imath] and [imath]s[/imath] symmetrically, since I assigned the same properties to both at each stage. Is this the correct formula for the problem? If not, then where have I gone wrong modeling this problem in terms of relation counting?
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