Update solution.md (#37)

Author: ToxicalNoob3062

src/content/questions/comp2804/2022-winter-final/15/solution.md

@@ -1,63 +1,17 @@
-Warning: I don't know but I can try
-
-<ul>
-	<li> Let S be the set of all possible permutations <br/>
-	$ |S| = 20! $
-	<li> Let's determine $A$ <br/>
-	We could pick 1 and any of the 19 larger integers and the other 18 numbers have 18! posible permutations <br/>
-	We could pick 2 and any of the 18 larger integers and the other 18 numbers have 18! posible permutations <br/>
-	\ldots <br/>
-	We could pick 19 and 20 and the other 18 numbers have 18! posible permutations <br/>
-	$ |A| = 19 \cdot 18! + 18 \cdot 18! + \ldots + 1 \cdot 18! $ <br/>
-	$ |A| = 18! \cdot \frac{(19)(20)}{2}$ <br/>
-	$ |A| = 18! \cdot (19)(10) $ <br/>
-	$ |A| = 18! \cdot 190 $
-	$ |A| = 10 \cdot 19! $
-	$ \Pr(A) = \frac{|A|}{|S|} $ <br/>
-	$ \Pr(A) = \frac{10 \cdot 19!}{20!} $ <br/>
-	$ \Pr(A) = \frac{10}{20} $ <br/>
-	$ \Pr(A) = \frac{1}{2} $
-	<li> Let's determine $\Pr(B)$ <br/>
-	We could pick 1 and any of the 19 larger integers and the other 18 numbers have 18! posible permutations <br/>
-	We could pick 2 and any of the 18 larger integers and the other 18 numbers have 18! posible permutations <br/>
-	\ldots <br/>
-	We could pick 19 and 20 and the other 18 numbers have 18! posible permutations <br/>
-	$ |A| = 19 \cdot 18! + 18 \cdot 18! + \ldots + 1 \cdot 18! $ <br/>
-	$ |A| = 18! \cdot \frac{(19)(20)}{2}$ <br/>
-	$ |A| = 18! \cdot (19)(10) $ <br/>
-	$ |A| = 18! \cdot 190 $
-	$ |A| = 10 \cdot 19! $
-	$ \Pr(A) = \frac{|A|}{|S|} $ <br/>
-	$ \Pr(A) = \frac{10 \cdot 19!}{20!} $ <br/>
-	$ \Pr(A) = \frac{10}{20} $ <br/>
-	$ \Pr(A) = \frac{1}{2} $
-	<li> Let's determine $\Pr(A \cap B)$ <br/>
-	Well, let's go I guess!
-	<ul>
-		<li> $\pi_{12}$ could be 1 and $\pi_{11}$ would have to be any of the 19 values smaller than it and $\pi_{10}$ would have to be any value larger than $\pi_{11}$. The remaining 17 values have 17! permutations
-		<li> $\pi_{12}$ could be 2 and $\pi_{11}$ would have to be any of the 18 values smaller than it and $\pi_{10}$ would have to be any value larger than $\pi_{11}$. The remaining 17 values have 17! permutations
-		<li> \ldots
-		<li> $\pi_{12}$ could be 19 and $\pi_{11}$ would have to be 20 and $\pi_{10}$ would have to be any value larger than $\pi_{11}$. The remaining 17 values have 17! permutations
-	</ul>
-	Idk if there's a way to express it but we can try <br/>
-	$ \sum_{i=1}^{18} ( \sum_{j=i+1}^{19} (\sum_{k=j+1}^{20} 1 \cdot 17!) ) $<br/>
-	$ 17! \cdot \sum_{i=1}^{18} ( \sum_{j=i+1}^{19} (\sum_{k=j+1}^{20} 1) ) $<br/>
-	$ 17! \cdot \sum_{i=1}^{20} ( \sum_{j=1}^{i-1} (\sum_{k=1}^{j-1} 1) ) $<br/>
-	$ 17! \cdot \sum_{i=1}^{20} ( \sum_{j=1}^{i-1} ( j - 1 ) ) $<br/>
-	$ 17! \cdot \sum_{i=1}^{20} ( \frac{(i-1)(i-2)}{2}) $<br/>
-	$ 17! \cdot \frac{1}{2} \cdot \sum_{i=1}^{20} ( (i-1)(i-2) ) $<br/>
-	$ 17! \cdot \frac{1}{2} \cdot \sum_{i=1}^{20} ( i^2 - 3i + 2 ) $<br/>
-	$ 17! \cdot \frac{1}{2} \cdot ( \sum_{i=1}^{20} i^2 - \sum_{i=1}^{20} 3i + \sum_{i=1}^{20} 2 ) $<br/>
-	$ 17! \cdot \frac{1}{2} \cdot ( \frac{20 \cdot 21 \cdot 41}{6} - 3 \sum_{i=1}^{20} i + 2 \sum_{i=1}^{20} 1 ) $<br/>
-	$ 17! \cdot \frac{1}{2} \cdot ( \frac{20 \cdot 21 \cdot 41}{6} - 3 \cdot \frac{20 \cdot 21}{2} + 2 \cdot 20 ) $<br/>
-	$ 17! \cdot \frac{1}{2} \cdot ( \frac{20 \cdot 7 \cdot 41}{2} - 3 \cdot \frac{10 \cdot 21}{1} + 40 ) $<br/>
-	$ 17! \cdot \frac{1}{2} \cdot ( \frac{10 \cdot 7 \cdot 41}{1} - 3 \cdot 10 \cdot 21 + 40 ) $<br/>
-	$ 17! \cdot \frac{1}{2} \cdot ( 10 \cdot 7 \cdot 41 - 3 \cdot 10 \cdot 21 + 40 ) $<br/>
-	$ 17! \cdot ( 5 \cdot 7 \cdot 41 - 3 \cdot 5 \cdot 21 + 20 ) $<br/>
-	$ 17! \cdot ( 5 \cdot 7 \cdot 41 - 3 \cdot 5 \cdot 21 + 20 ) $<br/>
-	$ 17! \cdot ( 1435 - 315 + 420 ) $<br/>
-	$ 17! \cdot ( 1540 ) $<br/>
-	ngl, the value is actually wrong. Basically the probability of A intersect B is smaller than the product of the 2 probabilities
-</ul>
-
-There's no way this is the actual way to do the question, right? This is bullshit. I am LITERALLY losing my sanity
+Here the thing is easy but tricky:
+
+For simplicity lets take a1 a2 and a3 are our permutations of set {1,2,3}
+
+so Pr(A=a1>a2) = 1/2 because swapping the values of the variables a1 and a2 will make the sign turn to > and violating the constraints. so we can swap in 2! ways and for all ways the signs will change except 1 way. ie : a1 > a2.
+
+under same logic we can say for Pr(B=a2>a3) = 1/2.
+
+Similarly for Pr(A intersect B ) = 1/3! = 1/6. because under only 1 condition a1>a2>a3 and swapping there values will lead to something else compared to the condition we want.
+
+so we can see Pr(A intersect B) < Pr(A) * Pr(B)
+
+what for the big numbers now ? like n = 100 or 200 instead of 3?
+
+actually n doesnt put an affect here because permutation is something know as a unform distributions.
+
+So irrespective of the value of n for a subset of k it will always be 1/k! as the probability to get a specific sign sequence.