Permutations and Combinations are CRAZY HARD. (1 Viewer)

[m.incognito]

I can't do it!

BAH. It's so difficult.
12 differently coloured beads are arranged around a necklace. How many different arrangements are possible?

The answer is 19 958 400 - i can't get it.

 

[morganforrest]

I remember we argued about that in class.....if you turn the necklace upside down its still the same so there are half as many possibilities

so to find arrangements in a necklace: [(n-1)!]/2

 

[m.incognito]

WHAAAAAAAAAAAAT!!!???

lol.

 

[morganforrest]

it shat me too until I got it

 

[JasonNg1025]

Your class learnt circle probability?

It's because in the arrangements in a line:

o--o--o--o--o...
^
First bead

But in a circle, there's no such thing as a first bead because all of them can be taken as first, so when you do 12!/2 it would be (12-1)!/2 = 11!/2 = your answer.

​

 

[Iheartpaulfrank]

Ohmygosh. I've been doing this in stats and no wonder I didn't get it. It was only covered in 3U Maths, it seems!

 

[Hikari Clover]

回复: Re: Permutations and Combinations are CRAZY HARD.

wait guys , why is 11!/2 ?

from what i learnt about circle probability, isnt it just 11!?

 

[shredinator]

Because the necklace can be viewed from either side:

What appears to be two different combos, is just one, because when you flip it they're the same.

Hard to explain without showing you.

But its just the circle thing: 11! divided by 2 because there are half as many possible combos.

 

[Hikari Clover]

回复: Re: Permutations and Combinations are CRAZY HARD.

but what if u just arrange 12 ppl in a circle, what's the difference?

 

[shredinator]

Well unless you get under the table, you're not going to be able to half the combos.

The point they're trying to make is if i have

............2
....7..............5

8......................1

....3...............4
............6


and you flip it over... its the same as :
...........2
...5..............7

1....................8

...4..............3
...........6

for each combo, there is another one that is indentical when you flip it, so there are only half the combos than there would be if you could only look at it in one direction.

 

[Hikari Clover]

回复: Re: Permutations and Combinations are CRAZY HARD.

i got it ~~~~~

so everytime when i see a necklace Q i should halve the answer:rofl:

 

[shredinator]

pretty much.

 

[Kujah]

Re: 回复: Re: Permutations and Combinations are CRAZY HARD.

:uhhuh:

 

[Hikari Clover]

回复: Re: Permutations and Combinations are CRAZY HARD.

man, i have a Q as well~~~

8 points r marked at random on a circle.how many triangles can be formed?

Answer:56

y is so simple like that just 8c3?

 

[Kujah]

Re: 回复: Re: Permutations and Combinations are CRAZY HARD.

Oh by the way, good luck guys

 

[morganforrest]

Re: 回复: Re: Permutations and Combinations are CRAZY HARD.

As Shredinator said, basically you can't turn a table of people on their heads but you can flip a necklace over to form the same combination

Morgan say: Concern yourself not with the why...only with the how

 

[iEdd]

Re: 回复: Re: Permutations and Combinations are CRAZY HARD.

but what if u just arrange 12 ppl in a circle, what's the difference?

Think of it this way. Beads are placed on a string along their axis, in which they appear identical every 180°. Humans at a table do NOT look the same from heads downwards as feet upwards, therefore you don't divide by 2.

 

[pLuvia]

Re: 回复: Re: Permutations and Combinations are CRAZY HARD.

man, i have a Q as well~~~

8 points r marked at random on a circle.how many triangles can be formed?

Answer:56

y is so simple like that just 8c3?

The reason is a triangle has three points and you can use any of the points on the circle to form a triangle. So you just pick three different points on the circle with 8 points so its just 8C3

 

[m.incognito]

Re: 回复: Re: Permutations and Combinations are CRAZY HARD.

okay i need more help.
god i hate this shit


A school committee is to be made up of 5 teachers, 4 students and 3 parents.
b) if Jan and her mother both apply find the probability that both will be chosen for the committee...


no. seriously. HUH?!

is it like... 4C1 x 3C1 or something? I HAVE NFI.

EDIT: rest of question:

12 teachers 25 students and 7 parents apply.

 

[Kujah]

Re: 回复: Re: Permutations and Combinations are CRAZY HARD.

How many teachers, students and parents all up?