combinatorics questions

can someone help me through these please.

for 1a) I got 9! which I think k is correct and for 1b) I got 9!-16, is there a proper method to do this because I just thought if it in my head. I'm unsure about 1c)

for 2a) I think that relation ab is symmetric but I don't know how to answer it...

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can someone help me through these please.

for 1a) I got 9! which I think k is correct and for 1b) I got 9!-16, is there a proper method to do this because I just thought if it in my head. I'm unsure about 1c)

1a is correct.

1b isn't.

Don't forget whilst Charlene is standing next to Anita, the others have numerous possibilities (for you to work out) of where to stand.

is it 9!-(9x8)? since if you were to pick 2 people without replacement and find all possibilities it would be 9x8.

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Nope.

These are not any two people. They are two specific people - Charlene and Anita. And you're looking to subtract the number of permutations of the 9 people where these two are together. It's considerably more than 9x8.

so in this case, im thinking that it'd be 9!-2(8!)?

Yep.

lol doing it the method I did doesn't help me in the next part though, I'm guessing that its 9!-2(8!)-4(6!)...
what is a 'rigorious' method to do this?

Don't know why you wrote -4(6!) there.

Adapting the method of part b is the way to go, though it becomes a little more complex. Think about what you're doing.

What did the 9!-2(8!) mean in part b, in english without the maths?

so in total there's 9! possible orders but since 1 and 2 do not want to stand next to each other, it reduces the shuffles of when they are standing next to each other, so say if we had 12xxxxxx and x12xxxxxx and so on moving them both to the right 8 times at Max, to reach xxxxxxx12.
for 12xxxxxxx it would be 9!-7! but this would happen 8 times so it would be 9!-8(7!). then we the situation of them standing like 21xxxxxxx so this would be another 8(7!) so in total it would be 9!-16(7!) which is the answer to that part since 16(7!)=2(8!)

I'm starting to realise that the method I done it in my previous post was BS.

OK so now thinking about the next part.

3 and 4 now don't wanna stand next to each other. i think its reduced by the same amount again but with an additional amount. so 1234xxxxx and these four can move along to the right 5 times so it would be minus 5(5!). 1234 can shuffle around like 4! times I think so its now 5(4!)(5!) so in grand total it'd be
9!-16(7!)-16(7!)- 5(4!)(5!)?

I doubt that's correct but that's all I can think of

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The way I think about it is. If you have two that are together, they can be ordered in 2! ways. This now gives us 8 items. So no. of orderings is 2(8!).



Couple of things.

Subtracting the 16(7!) twice - yep.

But you now need to cater for when the two pairs are both present. This will have been included in your initial subtractions twice! So you need to add.

The logic is the same as working out P(AuB), where A is the event the first pair are together and B the even the second pair are together.

And you know P(AuB) = P(A) + P(B) - P(AnB)

So, since you're subtracting AuB, you end up having to add AnB.

So, what are the number of permutations where we have 12 together and 34 together?

Note: This does not require 1234 all together.

answering what you asked about the permutations, it can be like 12xxxxx34. 12 and 34 can be shifiting in oppositing directions 3 times (whilst standing next to each other). but the pairs can switch sides so 34xxxxx12 so 2(3) we now have. and 12 and 34 can be 21 and 43 so its now 4(3)=12. since there are 5 left, permutations would be 5! but because of the shifting, it'd be 12(5!) permutations?

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