permutation + combinations help PLEASE! (2 Viewers)

[chicken4koo]

a dinner party is arranged for 16 people. the people will be seated along 2 sides of a rectangular table with 8 chairs on each side. 4 people wish to sit on one particular side and a couple wishes to sit together on the other side. how many ways can 16 people be arranged!!!!!

how do i get the answer to this? i'm lowkey stumped, i've got 3 different answers.......someone, anyone, help a poor girl out..,,,,

 

[ilovecutecats]

make sure the table is big enough

 

[chicken4koo]

make sure the table is big enough

i wish you all the worst on your future endeavours, love you girl!

 

[certified dummy]

this is straight off google cause im abhorred at perms and combs
some of them do say 10C4 * (8!)^2 using a different logic but its the same thing

personally id crash the tea party for the headache this gave me but thats the wonder of maths i suppose

 

[ShhQuietPlease]

I remember this question!

When doing permutations and combinations, a good technique is to sit people down one at a time and consider the number of options they have.

We first sit the picky people. Let's sit the four people (A, B, C, D) that want to sit on a particular side. On that particular side, there are initially 8 empty seats.
A: 8
B: 7
C: 6
D: 5

We then sit the couples (E, F). On the other side, there are initially 8 empty seats.
E: 8
F: 7

We then sit the rest of the people down. Since they can sit anywhere, then their options start with the number of empty chairs at the table (10).
G: 10
H: 9
I: 8
J: 7
K: 6
L: 5
M: 4
N: 3
O: 2
P: 1

We finally multiply the options together:
8 x 7 x 6 x 5 x 8 x 7 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 341397504000

 

[alewe]

I'm not sure if I'm correct, but I would do 10C4 x 8! x 7! x 2!, because you already know 4 people will be on one side, so then there are 10C4 ways to pick the other 4 people who will sit on that side. Then when you have those 8 people, the ways they can sit is calculated by 8!. Then on the other side of the table, I interpreted the question as asking for the couple to sit next to eachother, so you treat them as one unit, so the number of ways on the other side is 7!, the x 2! for the arrangements of the couple. I think @ShhQuietPlease did the calculations assuming the couple didn't have to sit together.

 

[r25nd]

i would tell the couple to suck it up and sit with the other 4

 

[C2H6O]

I'm not sure if I'm correct, but I would do 10C4 x 8! x 7! x 2!, because you already know 4 people will be on one side, so then there are 10C4 ways to pick the other 4 people who will sit on that side. Then when you have those 8 people, the ways they can sit is calculated by 8!. Then on the other side of the table, I interpreted the question as asking for the couple to sit next to eachother, so you treat them as one unit, so the number of ways on the other side is 7!, the x 2! for the arrangements of the couple. I think @ShhQuietPlease did the calculations assuming the couple didn't have to sit together.

I got the same answer, although my combs are very rusty

 

[SS173]

There is a good example in Cambridge Year 11 textbook on pp 674.

To solve this problem, let's break it down into steps.

1. Identify fixed placements:
   • Four specific people want to sit on one side of the table (let's call this Side A).
   • One couple wants to sit together on the other side (Side B).
2. Arrange the restrictions first:
   • For Side A, arrange 4 people in 8 seats : 8P4 (or 8C4x4!) = 1680
   • For Side B, as the couple sits together, there are only 7 blocks and they can switch : 7x2 = 14
3. Arrange the remaining 10 people:

There are 10 seats and 10 people remaining : 10!

• Total :

1680x14x10!