#multiplicative principle

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fickle drift
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When trying to find the cardinality of the set $\{(a,b,c): a,b,c\text{ are disjoint 2-element subsets of }{1,2,3,4,5,6}}$ I got $\binom{6}{2}\binom{4}{2}\binom{2}{2}$ but I don't really understand why it works. The multiplicative principle says that $\abs{A\times B\times C}=\abs{A}\abs{B}\abs{C}

cosmic lichenBOT
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magic kilnBOT
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Django
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fickle drift
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But what are sets A, B, and C here?

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Normally A is the set of possible values of a, B is the set of possible values of b and likewise for C

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But here B differs depending on A

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So it's confusing

fickle drift
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@upper matrix I do t get

upper matrix
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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

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well

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The idea is that no matter what a you choose you can always choose some b

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But I guess it’s double counting

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Wait no it’s not

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Because order matters

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That’s why you can just multiply them

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A,B,C are just sets of two elements

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So to visualize this

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For $A$, we can choose 2 distinct elements which there are 6 choose 2 ways to do

magic kilnBOT
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Magma (N, ^)#Magma4Manager

upper matrix
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For $B$ we can choose from the 4 that remain, that’s 4 choose 2

magic kilnBOT
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Magma (N, ^)#Magma4Manager

upper matrix
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And for $C$, well, we’re left with 1 possible way no matter what

magic kilnBOT
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Magma (N, ^)#Magma4Manager

upper matrix
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The fundamental idea is the number of ways we can choose B or C is independent of A

fickle drift
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So we need a multiplicative principle update

upper matrix
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While what we can choose from those is dependent on A, the count isn’t

upper matrix
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Which is why we’re allowed to multiply

fickle drift
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it's also 6!/2^3

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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}

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And u can probably figure out why we need to divide by 2^3

upper matrix
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yes

fickle drift
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And $\frac{(2n)!}{2^n}=\prod_{i=0}^{n-1}\binom{2n-2i}{2}$

magic kilnBOT
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Django

upper matrix
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Ah

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Django does it make sense now @fickle drift