Combinations help (1 Viewer)

[Dubble-u25]

Eight people attend a meeting. They are provided with two circular tables, one seating 3 people, the other 5 people.

(i) How many seating arrangements are possible?
(ii) If the seating is done randomly, what is the probability that a particular couple are on
different tables?

I don't have answers, so i'm not quite sure if I'm right :S
Thanks in advance

 

[muzeikchun852]

(i) 8C5 x (5-1)! x (3-1)! = 2688

8C5 - choose the people first.
(5-1)! x (3-1)! - arrange

(ii) 1-P(together)
1 - 7C5 x (5-1)! x (3-1)!/2688 <-- unsure.
=5/8

7C5 - treat the couple together.
(5-1)! x (3-1)! - arrange

 

[apollo1]

(i) 8C5
(ii) 2 x 6C4/8C5 <-- unsure.

why is it 8C5 for the first one isnt this a permutation (circular arrangement)?

 

[mimic]

i) In the table of 3, 1 person has to sit down as a starting point, so that means the remaining 2 seats can be filled by 7x6 people. The table of 5 then has to have one person as a starting point as well, leaving the 4 seats with 4x3x2x1 options. That means a total of 7x6x4x3x2x1= 1008 options.

Someone verify this^

ii) If the husband is on the table of 3, he has 3C1 options, leaving the wife with 5C1 options. 3C1 x 5C1= 15. If the couple swap tables, this is simply multiplied by 2, giving 30.

Therefore probability is equal to 30/1008 or 5/168.

Someone verify please^

 

[Kyrix]

i) In the table of 3, 1 person has to sit down as a starting point, so that means the remaining 2 seats can be filled by 7x6 people. The table of 5 then has to have one person as a starting point as well, leaving the 4 seats with 4x3x2x1 options. That means a total of 7x6x4x3x2x1= 1008 options.

Someone verify this^

ii) If the husband is on the table of 3, he has 3C1 options, leaving the wife with 5C1 options. 3C1 x 5C1= 15. If the couple swap tables, this is simply multiplied by 2, giving 30.

Therefore probability is equal to 30/1008 or 5/168.

Someone verify please^

(i) Well, i dont think it works like that.

If you place one person at table A and then place one person at table B, as place markers, then the rest of the 6 people can be sitted in 6! ways

Place person at table A = 1
Place person at table B = 1
Left over people have a arrangement = 6x5x4x3x2x1
By Mutilplication Principle No. of arrangments = 6! x 1 x1 = 720

(ii) I think your right about two, but the sample space should be 720
so 30/ 720 = 1/24

 

[someth1ng]

This question was in my half-yearly exam - or a very similar one.

To do this one:
a. A=8C5 x (5-1)! x (3-1)!=2688 ways

8C5 is used to pick 5 from the 8 people to go into the 5 member table. 8C3 can be used instead of the 8C5 as they are the same thing. Both 8C3 and 8C5 must not be multiplied together as if one combination is picked, the other table is automatically chosen as the ones not picked.
(5-1)! for circular arrangement around the 5 member table.
(3-1)! for circular arrangement around the 3 member table.

b. P(different tables)=[6C4 x (5-1)! x (3-1)! x 2]/A=1440/2688=15/28

6C4 is to place the remaining 6 people after assuming each is on a different table.

(5-1)! is to arrange the 5 people around the table.
(3-1)! is to arrange the 3 people around the table.
Arranging these 5 and 3 people are including the couple.
Multiplying this by 2 to find the ways when they are swapped tables (the couple).

Place person at table A = 1
Place person at table B = 1
Left over people have a arrangement = 6x5x4x3x2x1
By Mutilplication Principle No. of arrangments = 6! x 1 x1 = 720

This makes no sense at all:
1. You have forgotten the second table.
2. You need to pick people to join which table. ie. how many combinations of table members can be made.

 

[Kyrix]

This question was in my half-yearly exam - or a very similar one.

To do this one:
a. A=8C5 x (5-1)! x (3-1)!=2688 ways

8C5 is used to pick 5 from the 8 people to go into the 5 member table. 8C3 can be used instead of the 8C5 as they are the same thing. Both 8C3 and 8C5 must not be multiplied together as if one combination is picked, the other table is automatically chosen as the ones not picked.
(5-1)! for circular arrangement around the 5 member table.
(3-1)! for circular arrangement around the 3 member table.

b. P(different tables)=[6C4 x (5-1)! x (3-1)! x 2]/A=1440/2688=15/28

6C4 is to place the remaining 6 people after assuming each is on a different table.

(5-1)! is to arrange the 5 people around the table.
(3-1)! is to arrange the 3 people around the table.
Arranging these 5 and 3 people are including the couple.
Multiplying this by 2 to find the ways when they are swapped tables (the couple).


This makes no sense at all:
1. You have forgotten the second table.
2. You need to pick people to join which table. ie. how many combinations of table members can be made.

Hmm yes
Youre probly right.
Im very weak at combinations, sorry for any confusion I could of caused =/

 

[D94]

To do this one:
a. A=8C5 x (5-1)! x (3-1)!=2688 ways
...
b. P(different tables)=[6C4 x (5-1)! x (3-1)! x 2]/A=1440/2688=15/28
...

^ This is correct.

Another way you could look at the second part in these types of questions is to look at the probability for one person to sit on one table, and the other person to be sitting on the other table: = 3/8*5/7 + 5/8*3/7 = 15/28

3/8 = person A can choose to sit on the table of 3 and has 3 places out of a total of 8 places.
5/7 = person B must choose to sit on the other table of 5 so he/she can sit 5 places out of a total of 7; 7 because person A has already taken one place
+ (plus) = OR situation, as the arrangement can be down considering the table of 3 as the first table or the table of 5 as the first table.
5/8 = as above.
3/7 = as above.

 

[someth1ng]

Another way you could look at the second part in these types of questions is to look at the probability for one person to sit on one table, and the other person to be sitting on the other table: = 3/8*5/7 + 5/8*3/7 = 15/28

I've actually never thought of doing it like that, it's very neat!

Im very weak at combinations, sorry for any confusion I could of caused =/

It's okay, just do some practice and understand the concept of P&C. I recommend going on YouTube and searching "Permutations and Combinations" as it really helped me figure out how to do it all.

 

[Vidhya]

This was a good question! Perms and Combs are the only things I'm able to do confidently haha, so hopefully there's a big chunk of it in the exam :3

 

[someth1ng]

This was a good question! Perms and Combs are the only things I'm able to do confidently haha, so hopefully there's a big chunk of it in the exam :3

A "big chunk" for P&C would be about...6 marks worth. LOL

 

[Kyrix]

I recommend going on YouTube and searching "Permutations and Combinations" as it really helped me figure out how to do it all.

Could you direct me to a particular video?

All I could find were people teaching the basics

 

[someth1ng]

Could you direct me to a particular video?

All I could find were people teaching the basics

The Khan Academy ones were quite good. There's 3 videos AFAIK and in P&C, you generally only need to know the basics as that's all you are usually asked for. I don't remember seeing a HSC P&C questions that was extremely hard but rather simple 1 (sometimes 2) mark questions.

 

[Dubble-u25]

Thanks guys
I hate perm and com they're so hard sometimes

 

[someth1ng]

Thanks guys
I hate perm and com they're so hard sometimes

NP, P&C is generally quite simple in the HSC exam and are always worth only afew marks (typically 2-6 marks in total).