We can count the total choices of 3 scoops of ice cream chosen from 23 flavour (with repeat flavours allowed) by counting the arrangements of 22 bars | and 3 scoops ●.
From basic combinatorics the number of arrangements of n identical objects of one type with m identical objects of another type is (n+m)!/(n!m!). In this case n=22, m=3 so
(22+3)!/(22!3!) = 25!/(23!3!) = 2300 <----
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this can also be written as the binomial coefficient C(22+3,3) = C(25,3).
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There are 23 categories separated by 22 bars | and there are 3 scoops ● that can be placed in any of the categories e.g.
● ● ● | | | | | | | | | | | | | | | | | | | | | | (3 scoops of flavour 1)
● ● | ● | | | | | | | | | | | | | | | | | | | | | (2 scoops of flavour 1 and 1 scoop of flavour 2)
● | | | | | | | | | | | | | ● | | | | | | | | | ● (1 scoop each of flavours 1, 14 and 23)
| | | | | | | | | | | | | | | | | | | ● | | ● ● | (1 scoop of flavour 20 and 2 scoops of flavour 22)
etc.
We can count the total choices of 3 scoops of ice cream chosen from 23 flavour (with repeat flavours allowed) by counting the arrangements of 22 bars | and 3 scoops ●.
From basic combinatorics the number of arrangements of n identical objects of one type with m identical objects of another type is (n+m)!/(n!m!). In this case n=22, m=3 so
(22+3)!/(22!3!) = 25!/(23!3!) = 2300 <----
---------------------
this can also be written as the binomial coefficient C(22+3,3) = C(25,3).
By different 3 scoop do you mean you can't have say, triple scoop of vanilla? If so then you can order 23*22*21 = 10626 different 3-scoop dishes.
If however you include say, triple scoop of vanilla as one of the possible 3-scoop dishes then you can have 23^3 = 12167 different 3-scoop dishes.
Lots
23(22)(21) calculate
I don't know