perms and combs: circular arrangements involving groups (2 Viewers)

[kractus]

eight people sit around a table. If the eight people consist of four sets of twins, how many arrangements are possible if each twin must sit opposite each other?

 

[mmmmmmmmaaaaaaa]

For this, you should be able to set each set of twins as one unit, so there are now four units (n=4) to set around the table. In this case as it is as they are set out in a circle, we can fix one of the units in place --> number of arrangements in a circle is (n−1)!,
Where n = number of arrangements

That is (4-1)! = 3!

However, since in each unit there are 2 people and can switch places, this doubles the number of arrangements.

ie. 2^n

Therefore, the solution would be 3! x 2^3 = 48

A bit rusty on perms and combs, hopefully this is right

 

[kractus]

For this, you should be able to set each set of twins as one unit, so there are now four units (n=4) to set around the table. In this case as it is as they are set out in a circle, we can fix one of the units in place --> number of arrangements in a circle is (n−1)!,
Where n = number of arrangements

That is (4-1)! = 3!

However, since in each unit there are 2 people and can switch places, this doubles the number of arrangements.

ie. 2^n

Therefore, the solution would be 3! x 2^4 = 96



A bit rusty on perms and combs, hopefully this is right

this is exactly how i did it xD but the answer is 48 so im pretty confused still thank u though!

 

[kractus]

also with this i tried 5C1 / 5C4 * 4! but apparently its wrong cause the answer is 125

and this idk where to start tbh

 

[carrotsss]

this is exactly how i did it xD but the answer is 48 so im pretty confused still thank u though!

You have to multiply it by 2^3, not 2^4 because it’s a circle, and so one of the twin groups is already fixed and can’t be swapped

 

[carrotsss]

View attachment 38290
also with this i tried 5C1 / 5C4 * 4! but apparently its wrong cause the answer is 125
View attachment 38291
and this idk where to start tbh

First question:
There are four students, so let’s go through each of them in order. The first student can go into any house. The second student now needs to go into that house as well, so that’s 5. Then the third student also needs to get into that house, so that’s another 5, and then the fourth students needs to get in which is another 5.
Hence, we do 1(first)*5(second)*5(third)*5(fourth)=125

The second question is a bit weird since usually they limit it to “winning in 4 rounds” but I’ll give it a go once I’ve got pen and paper

 

[hscgirl]

View attachment 38291
and this idk where to start tbh

is the answer to the second question 6/11?

let the probability of katherine winning be 'x'
1 = x + (5/6)x
^^because 5/6 is the probability of katherine not rolling a six on the first go, so andrew's probability of winning on the next go is now 'x', so multiplying them together is (5/6)x. also we are adding (5/6)x and x together to equal 1, as one of the events are certain to happen
by rearranging the equation, x = 6/11

idk if im right lol but hopefully i am

 

[Average Boreduser]

bro I hate perms and combs with a mfing pashion

 

[gazzaboy]

eight people sit around a table. If the eight people consist of four sets of twins, how many arrangements are possible if each twin must sit opposite each other?

A couple of other ways to think about it (I find that it always helps to have multiple ways)

A. Another way of thinking about anything around a circle is "there is only one first choice" which then fixes the circle.

• So place any person in the circle (the "first person"): 1 way. This fixes their twin opposite: 1 way
• You now have 6 spots left, so there's 6 ways to place the next person, say, to the left of the first person. This automatically fixes their twin opposite of them.
• Now there are 4 spots left, so and there are 4 ways to place the next person, say 2 spots to the left of the first person. Again, this fixes their twin opposite.
• Lastly, there are 2 spots left, and there are 2 ways to place the next person 3 spots to the left of the first person (it could be either of the twins).

So the number of ways is 1 * 1 * 6 * 4 * 2 = 48.

B. Figure out how many ways there are to arrange them in a line such the twins are four spots away, and then divide by 8 (because each "line arrangement" is the "same" as 7 other "line arrangements" if considered in a circle). Imagine there are 8 spots in a line.

• Place a person in the first spot: 8 ways -> this automatically pins down the 5th spot as the twin
• Place a person in the second spot: 6 ways ->this automatically pins down the 6th spot as the twin
• Place a person in the third spot: 4 ways -> this automatically pins down the 7th spot as the twin
• Place a person in the fourth spot: 2 ways -> this automatically pins down the 8th spot as the twin

Hence the number of ways to arrange in a line is 8 * 6 * 4 * 2 = 384, so to get to way to arrange in a circle, we divide by 8 to get 48.

 

[gazzaboy]

View attachment 38290
also with this i tried 5C1 / 5C4 * 4! but apparently its wrong cause the answer is 125
View attachment 38291
and this idk where to start tbh

First one: carrotsss's answer is good. You can also do it using thinking about it with your standard probability rules i.e.
Pr(same house) = Pr(all in house 1) + Pr(all in house 2) + Pr(all in house 3) + Pr(all in house 4) + Pr(all in house 5) = 5 * Pr(all in a particular house) = 5 * 1/5 * 1/5 * 1/5 * 1/5 = 1/125

Second one: hscgirl's answer is brilliant but can be a bit tricky to see. If x = Pr(Katherine wins), then we know that 1 = Pr(Katherine wins) + Pr(Andrew wins) (since either one of them will win. But there's a 5/6 chance that Katherine doesn't win in the first go, so Pr(Andrew wins) = 5/6 * Pr(Katherine wins).

Another way is to break it down into the different ways Katherine can win...

Pr(Katherine wins) = Pr(Katherine wins in the first roll) + Pr(Katherine wins in her second roll) + Pr(Katherine wins in her third roll) + ...

It turns out this is an infinite series!

Pr(Katherine wins in the first roll) = 1/6
Pr(Katherine wins in her second roll) = 5/6 * 5/6 * 1/6 (since Katherine and Andrew have to both not roll a 6)
Pr(Katherine wins in her third roll) = 5/6 * 5/6 * 5/6 * 5/6 * 1/6
... and so on.

Summing it up you have a geometric series where the first term is 1/6 and r = 5/6 * 5/6 = 25/36. Hence

 

[SMARTYPANTS_SINGH]

bro I hate perms and combs with a mfing pashion

I hate it so much that I made a whole thread on it