pls help with perm comb q (1 Viewer)

[Plshelpmewithmaths]

10 students in a circular table, number of arrangments if 3 particualr students cant sit next to each other.

 

[cossine]

10 students in a circular table, number of arrangments if 3 particualr students cant sit next to each other.

Just consider the different cases

 

[Run hard@thehsc]

I tried using complementary. That is 9! - ((8-1)! *3!). I am not sure if this could be right. Can someone correct me if I am wrong? Thanks

 

[Como90]

I tried using complementary. That is 9! - ((8-1)! *3!). I am not sure if this could be right. Can someone correct me if I am wrong? Thanks

i think you also have to consider cases where only two sit next to each other

 

[Run hard@thehsc]

i think you also have to consider cases where only two sit next to each other

oh yh fax. Then it would be 9! - ((8-1)! * 3!) - ((9-1)! * 2! *3C2) --> feel like I butchered this lol...

 

[Como90]

i think you can first arrange 7 people then insert and arrange the 3 people in 7 gaps between the 7 people which gives
(7-1)! * 7P3 ?

 

[carrotsss]

My working (no clue if it’s correct, my perms/combs are pretty rusty):

- Fix one of the restricted people at the top of the table, then choose 2 out of 7 seven unrestricted people to sit next to them (7P2)
- 7! spots left for everybody else to sit in, except there are 12 of these possibilities where the two remaining restricted people sit next to each other so it’s (7!-12)

So my final answer is 7P2*(7!-12)

… which gives a different answer to @Como90, who I think is correct, so idk where I went wrong

 

[Como90]

My working (no clue if it’s correct, my perms/combs are pretty rusty):

- Fix one of the restricted people at the top of the table, then choose 2 out of 7 seven unrestricted people to sit next to them (7P2)
- 7! spots left for everybody else to sit in, except there are 12 of these possibilities where the two remaining restricted people sit next to each other so it’s (7!-12)

So my final answer is 7P2*(7!-12)

… which gives a different answer to @Como90, who I think is correct, so idk where I went wrong

you have to arrange everyone else with those 12 possibilities
you should instead get:
7P2 *(7! - 2!*6!) which gives my answer

 

[tgone]

I also get the same answer, but I consider the complement as well;

case 1, all three students together

ways=7! * 3!

case 2, two students together


ways=(3C2*2!)(7C2*2!)(6!)=3P2*7P2*6!

[i prefer the ‘choose’ operator as it makes more intuitive sense to me, they’re identical though]

then the complement is 9! - (case 1 + case 2), which computes to the same as Como

 

[Plshelpmewithmaths]

i think you can first arrange 7 people then insert and arrange the 3 people in 7 gaps between the 7 people which gives
(7-1)! * 7P3 ?

i did 6! x 7C3 instead of 7P3, cuz ig it doesnt matter where they go?

 

[tgone]

i did 6! x 7C3 instead of 7P3, cuz ig it doesnt matter where they go?

it does matter, because there are different arrangements of the 3 depending on which spots are chosen, 7C3 chooses the spots but doesn't order the 3 within them

 

[Deem_Skills]

this question becomes easier because the restriction is that the 3 people cant sit together. You would likely have to use comp events if it was "3 people can't ALL be seated in a group"

the best way to approach this q is by considering the number of ways these 3 people are separated. First, arrange the others --- > (7-1)!, then place the 3 people that can't sit together in between the 7 people. ----- > 7P3. The 7P3 comes from the fact that there are 7 possible seats in between the 7 people we placed, so the ways of placing the 3 people are 7 * 6 * 5 = 7P3. If you were to use 7C3, you must think of it like this: 7C3 will only CHOOSE the 3 seats for the 3 people to go into, so on top op the 7C3, we also need to multiply it by the ways they can be arranged in the 3 seats, so that would be 7C3 * 3!

So total ways = 6! * 7P3 or 6! * 7C3 * 3!