Permutations

[median breach]

3 green, 2 red and one blue are arranged, if we considered both of arrangements below to be the same then we will only be left with 30 different permutations.

6!/3!×2!×2!

Out of those 30, how many permutations will the two red beads be next to each other?

If anyone has any good resources to explain permutations, I would like to check them

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[soft quail]

You can treat the red beads as one object and count the permutations that way.

Basically, forcing them to be together.

[trail pasture]

As Kocher mentioned, 2 red beads can be viewed as one block. Then we will have 5!

We need to then divide by the repetitions of green so 5!/3! which is 20

Note that we don't divide by repetitions of the 2 red beads as they were considered to be one block earlier

[trail pasture]

On yt there are not many good vids explaining permutations and combinations but I recommend looking at the cambridge past papers for A level which have some good questions and there are yt vids explaining those

[burnt idol]

@median breach

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[median breach]

@trail pasture @soft quail thank you both so much

[lilac sparrow]

dont we need to divide by 2 cuz reflections are the same permutations??

cuz they're in that 30 permutations

I'm pretty sure its 10 and not 20.

[trail pasture]

wdym reflections?

where did the 2 come from

[lilac sparrow]

Identical reflections

[lilac sparrow]

like the pic

[soft quail]

Good eye.

[trail pasture]

I dont think I hv studied this yet

so does it narrow down the arrangements?

how does it work in this case?

[soft quail]

Like, it depends if you treat the beads independentely.

[trail pasture]

so like distinct?

[soft quail]

Yeah.

[trail pasture]

oh

[soft quail]

Just call the beads G_1 and G_2 and see what happens.

[trail pasture]

I was assuming that the beads were not distinct

[trail pasture]

ah yea that makes sense