Permutations and Combinations (1 Viewer)

Joined
Feb 16, 2014
Messages
2,258
Gender
Male
HSC
2014
Find the number of ways in which 4 girls and 3 boys can be seated in a row so that no two boys are next to each other.

Need confirmation to my answer.
 

Squar3root

realest nigga
Joined
Jun 10, 2012
Messages
4,994
Location
phenchod
Gender
Male
HSC
1998
Uni Grad
2005
not sure if correct tho.

= all combinations - boys sitting together
= 4!*3! - 5!/3!
= 124

must be a 4U question lol. I think in 3U the hardest it every gets is 2 boys and then you do the usual
 

Squar3root

realest nigga
Joined
Jun 10, 2012
Messages
4,994
Location
phenchod
Gender
Male
HSC
1998
Uni Grad
2005
ohh wait, the only possible arrangement is

G B G B G B G

= 4!*3! = 144

so answer = 144?
 

panda15

Alligator Navigator
Joined
Feb 22, 2012
Messages
675
Gender
Male
HSC
2013
ohh wait, the only possible arrangement is

G B G B G B G

= 4!*3! = 144

so answer = 144?
G B G B G B G
B G B G B G G
B G G B G B G
B G B G G B G
G G B G B G B

So I think it would be 5*144?

I dunno. I hated perms and combs.
 

Squar3root

realest nigga
Joined
Jun 10, 2012
Messages
4,994
Location
phenchod
Gender
Male
HSC
1998
Uni Grad
2005
G B G B G B G (1)
B G B G B G G (2)
B G G B G B G (3)
B G B G G B G (4)
G G B G B G B (5)

So I think it would be 5*144?

I dunno. I hated perms and combs.
ohh i didn't think of those cases (lol square missed 80% :/)

but yeah that seems "more right"

EDIT:
[soz for labeling your post]

but aren't cases 2 & 4 "the same" because they are just "shifted" likewise all the others?
 
Last edited:
Joined
Feb 16, 2014
Messages
2,258
Gender
Male
HSC
2014
ohh wait, the only possible arrangement is

G B G B G B G

= 4!*3! = 144

so answer = 144?
you can have G G B G B G B or B G G B G B G, there's more arrangements to not allow two boys to be together. It just seems ambiguous and the way i did it seems dodgy
 
Joined
Feb 16, 2014
Messages
2,258
Gender
Male
HSC
2014
What i did was total combinations so 7! - (3x6!x2)

There are three boys, and total number of ways arranging two of them together is 3x6!x2!
 

panda15

Alligator Navigator
Joined
Feb 22, 2012
Messages
675
Gender
Male
HSC
2013
ohh i didn't think of those cases (lol square missed 80% :/)

but yeah that seems "more right"

EDIT:
[soz for labeling your post]

but aren't cases 2 & 4 "the same" because they are just "shifted" likewise all the others?
They're different because it's in a row. in (2) the 5th seat is a boy, in (4) it's a girl which is a different combination for the whole row.
 

Squar3root

realest nigga
Joined
Jun 10, 2012
Messages
4,994
Location
phenchod
Gender
Male
HSC
1998
Uni Grad
2005
They're different because it's in a row. in (2) the 5th seat is a boy, in (4) it's a girl which is a different combination for the whole row.
ohh yes because it is a permutations not a combination so the order is important.

soz, engrish is not my strong suite
 

dunjaaa

Active Member
Joined
Oct 10, 2012
Messages
473
Gender
Male
HSC
2014
This question involves the inclusion-exclusion principle. What I typically do for these types of questions is use spaces. So sit the girls down and leave spaces in between. E.g. _G(1)_G(2)_G(3)_G(4)_. Now notice how there's 5 potential spots that the boys can sit since the question says no two boys can sit together which implies that they are separated by all the girls. Out of the 5 spots, we choose 3 boys. But bear in mind the boys can be arranged in 3! ways and the girls in 4! ways. So by the multiplication principle, the total number of arrangements is 5C3 x 3! x 4! = 1440 ways
 

Squar3root

realest nigga
Joined
Jun 10, 2012
Messages
4,994
Location
phenchod
Gender
Male
HSC
1998
Uni Grad
2005
This question involves the inclusion-exclusion principle. What I typically do for these types of questions is use spaces. So sit the girls down and leave spaces in between. E.g. _G(1)_G(2)_G(3)_G(4)_. Now notice how there's 5 potential spots that the boys can sit since the question says no two boys can sit together which implies that they are separated by all the girls. Out of the 5 spots, we choose 3 boys. But bear in mind the boys can be arranged in 3! ways and the girls in 4! ways. So by the multiplication principle, the total number of arrangements is 5C3 x 3! x 4! = 1440 ways
Are you just making up principles?

Notsureifsrs
 

braintic

Well-Known Member
Joined
Jan 20, 2011
Messages
2,137
Gender
Undisclosed
HSC
N/A
This question involves the inclusion-exclusion principle. What I typically do for these types of questions is use spaces. So sit the girls down and leave spaces in between. E.g. _G(1)_G(2)_G(3)_G(4)_. Now notice how there's 5 potential spots that the boys can sit since the question says no two boys can sit together which implies that they are separated by all the girls. Out of the 5 spots, we choose 3 boys. But bear in mind the boys can be arranged in 3! ways and the girls in 4! ways. So by the multiplication principle, the total number of arrangements is 5C3 x 3! x 4! = 1440 ways
Correct working, but this is not really the inclusion-exclusion principle (although that is a way of answering this question).
Inclusion-exclusion involves subtracting/adding intersections of sets, based on a Venn diagram.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top