#multiplicative principle
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But what are sets A, B, and C here?
Normally A is the set of possible values of a, B is the set of possible values of b and likewise for C
But here B differs depending on A
So it's confusing
@upper matrix I do t get
Well you can chose 2 from 6 for the first two subsets, then you’ll have 4 left, so you can chose 2 from 4, and the remaining 2 chose 2 subset
well
The idea is that no matter what a you choose you can always choose some b
But I guess it’s double counting
Wait no it’s not
Because order matters
That’s why you can just multiply them
A,B,C are just sets of two elements
So to visualize this
For $A$, we can choose 2 distinct elements which there are 6 choose 2 ways to do
Magma (N, ^)#Magma4Manager
For $B$ we can choose from the 4 that remain, that’s 4 choose 2
Magma (N, ^)#Magma4Manager
And for $C$, well, we’re left with 1 possible way no matter what
Magma (N, ^)#Magma4Manager
The fundamental idea is the number of ways we can choose B or C is independent of A
So we need a multiplicative principle update
While what we can choose from those is dependent on A, the count isn’t
Yeah
Which is why we’re allowed to multiply
it's also 6!/2^3
Take a permutation of 1,2,3,4,5,6 (a1,a2,a3,a4,a5,a6) make a={a1,a2}, b={a3,a4}, c={a5,a6}
And u can probably figure out why we need to divide by 2^3
yes
And $\frac{(2n)!}{2^n}=\prod_{i=0}^{n-1}\binom{2n-2i}{2}$
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