#COMBINATIONS IN EVERYTHING

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old garnet
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@warped stone i wanna learn combinations so i can use them in anything. can u teach me the logic

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lusty sunBOT
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warped stone
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first one you take 13 choose 9

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basically, 13C9

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if you know the notation nCk = n!/k!(n-k)! you are good

old garnet
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yupyup

warped stone
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for 4 of the empty seat next to eachother

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you can just make it a sort of block of 4 empty row

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wait

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you dealing with arrangements

old garnet
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yes

warped stone
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which is nAk or n!/(n-k)!

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now

old garnet
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its permutations but i dont want to solve it using that i wanna use C or factorials

warped stone
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it is just 9!.13C4

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so yeah pretty much

old garnet
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oo

warped stone
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9! for different 9 arrangements and 13C4 for choosing 4 out 13 row to be empty

old garnet
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ooo

warped stone
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the second problem is solved by condensed the empty 4-row into a block of empty row

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as you see it have to be next with eachother so might as well make it into a block

old garnet
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oo yesyes

warped stone
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the you be having 10 row

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with one of them is the empty block

old garnet
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so 10!

warped stone
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so it have 10 choice of row

old garnet
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oo

warped stone
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and 9 permutation for the rest which is 9! which is 10.9! so you are correct

old garnet
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ohh

warped stone
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and since we have the probability problem as last

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consider this

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the quickest way to do this

old garnet
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i take out the arrangements of when they are not together and divide it by total?

warped stone
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is to take the number of arrangements for 9 cars in any position, and minus the number arrangements where the there is this empty block of row

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the reasoning here is that you dont wanna pick the one with a block of empty row

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but the other is fine.

old garnet
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that gives me when they are not togetehr right?

warped stone
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so you just subtract the one you dont wanna have to have a set of only the thing that you want.

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yeah

old garnet
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oo

warped stone
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like you have 10 balls, 5 blue, 5 red, you want red so you subtract the blue, you now left with 5 balls which is red.

old garnet
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ooo

warped stone
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and so just like the permutation

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and we can use our last result to subtract with eachother to have the final answer

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which is 9!.13C4 - 10!

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divide to 9!.13C4

old garnet
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ooo

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thanku!

warped stone
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np

old garnet
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and 1 more question

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when and why do we divide by number of repititions

warped stone
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well there are a reason for thing like combinations and permutation, arrangements

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permutation is like you have n object

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move it around.

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you will have n! possible position.

old garnet
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oo

warped stone
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arrangements is like you have n object. But say you just want k of those n object, like you want for example to take 2 balls from a set 4 balls, you pick your first object, then your second, till you are satify.

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like i pick one ball then another ball.

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for n object first pick you have n choice.

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second, one less than that cause one is on your hand.

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the third is one less than that again cause now two is on your hand.

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and you pick till you are satisfy

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So there are n.(n-1).(n-2)...(n-k-1) choice which is n!/(n-k)!

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so like permutation, but if you only want a smaller amount of object.

old garnet
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oo

warped stone
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combination is different. as it order dont matter to it.

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the permutation and arrangement will count the different position of the object.

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so for it abc is different than acb so it will count it as like it is two different object

old garnet
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oo

warped stone
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but combination

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dont

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they see abc and acb or whatever the next position is they be having is the same

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it only counting the elements

old garnet
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oo

warped stone
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abc is the same as acb for the combination

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but abc and acd is not.

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the way they count combination came from one key insight.

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you want to choose k objects, and you didnt care which order they came in.

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but for any k object of the same elements

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there of course k! possible position it can be in.

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but mathematician notice that, "well one only way to fix that, is to make all of it into a single object"

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by divide the arrangements of k object to k!

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since for example

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abc acb cab cba bca bac have 6 as they have 3! = 3.2.1 = 6 way to position themself

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to count them as one

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we just divide it.

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to 6

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so now every 6 element of it became one.

old garnet
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ohhhhhhh

warped stone
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generalized this idea give nCk

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or nAk/k!

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or n!/(n-k)!k!

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so permutation for when you want the position of the full thing

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arrangement when you just want the every permutation and possible way you can have k object from n object

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and combination when you just want the combination of k object, no order.

old garnet
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BRO thanku so much

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i have been confused for weeks now this clears everything up

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+clos

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+close