+88 A. In how many ways can five red chairs and three white chairs be arranged such that no two white chairs are next to each other?
B. In how many ways can four red chairs, three white chairs, and two blue chairs be arranged such that there is at least one red and one blue chair between two white chairs?
For A, there are initially 8!/5!3! = 5040 ways to arrange the chairs. Using complementary counting will be easier, so I just counted the number of ways that at least two white chairs are next to each other. There are 7 ways for 2 white chairs to be next to each other and 6 ways for 3 white chairs to be next to each other. Subtract this from 5040 to get 5027. Could somebody check this?
As for B, I'm not sure how to count the number of ways.
thank you in advance!
+129899 Here's what I get
A) Assuming that the chairs of each color are identical
You have calculated the the number of possible arrangements correctly
If two white chairs are together they could occupy positions
1,2 and the other white chair can occupy positions 4,5,6,7 or 8
2,3 and the other white chair can occupy positions 5,6,7 or 8
3,4 and the other white chair can occupy positions 1, 6, 7 or 8
4,5 and the other white chair can occupy positions 1,2, 7 or 8
5,6 and the other white chair can occupy positions 1,2,3 or 8
6,7 and the othe white chair can occupy positions 1,2,3 or 4
7,8 and the other white chair can occupy positions 1,2,3,4,5
So.....we have 30 arrangements here
Three white chairs together can occupy any of 6 positions
So......the total number of excluded arragements = 36
So....the total number of acceptable arrangememts = 5040 - 36 = 5004
+118687 Here is my take.
A
In how many ways can five red chairs and three white chairs be arranged such that no two white chairs are next to each other?
I am also assumiong that all the white chairs are identical and all the red chairs are identical.
And I am assuming that the chairs are placed in a row
The white chairs must be seperated so I can start with
W R W R W
Now there are 3 more red ones so I will put hashes where one or more of them can go
# W # R W # R W #
If all three are together then there are 4 places they can go
If 2 are together and the other seperate then there are 3+3+3+3=12 ways
If none are placed together then we have 4C3=4ways
4+12+4= 20 ways
20 ways is my answer
B
B. In how many ways can four red chairs, three white chairs, and two blue chairs be arranged such that there is at least one red and one blue chair between two white chairs?
The white chairs are seperated by at lease 1 red and one blue chair so there is the start
W [BR] W [BR] W
Of course the BR could also be RB and that goes for the second lot of BR as well so there are 4 copies of all continuing arrangements
(I could be double overcounting but I don't think so)
So
W [BR] W [BR] W Ive used all the White and all the blue chairs already there are only 2 red ones unused so where can I put them?
# W #[BR]# W #[BR]# W #
If they both stay together then there are 6 places they can go
If they are seperated then there are 6C2 = 15 places they can go.
Which adds to 15+6=21 places
But there were 3 other starter arrangments
4*21 = 84 possible arrangments
+118687 We will expect a response from you grs75.
Preferably with the answer you are given as correct.
So far you do not have flawless history of showing appreciation.
https://web2.0calc.com/questions/question-on-polynomial-division-question#r2
+88 I'm sorry. I was expecting to receive an email notification but did not get one, and I am not on this website 24/7. I did not intend to come off as unappreciative.
As for the correct answers, the answer to part A is 20. A pair of white chairs must be separated by one red chair. The red chairs can be arranged like _ R _ R _ R _ R _ R _ with 6 spaces to put the white chairs. Then 6 choose 3 = 20. Part B is not so clear to me, so I'm not sure of the correct answer.
+118687 Thanks Grs75.
Your response is appreciated.
It seems my part A and your given answer are the same anyway.
That is a good start :)
+118687 ok thanks for that, I thought I might have double counted.
Ok here is the go
the first bit of what I did was correct
# W #[BR]# W #[BR]# W #
If they both stay together then there are 6 places they can go
If they are seperated then there are 6C2 = 15 places they can go.
Which adds to 15+6=21 places
now lets look at a scenario
consider starting with
# W #[BR]# W #[RB]# W #
If I added 2 reds tied together they will all be different than any permutation in the first set, that is 6 ways
If 2 two added reds are to be added seperately then
# W #[BR]# W R[BR]# W #
will all be the same as
# W #[BR]# W #[RB]R W #
and there are 4 places where the second R can be put, so if I add15 in this second set I will have double counted by 4
So rather than 15 there will be 11
The next set will have 4 less again and the same with the last set
So rather than it being
21+21+21+21
it will be
21+17+13+9 = 60
+2668 Here's another way for 1 using Complementary Counting:
There are \({8 \choose 5} = 56\) ways to order them.
If only 2 white chairs are together, there are 7 positions they can occupy (1 and 2, 2 and 3, 3 and 4, and so on)
If the white chairs occupy spots (1 and 2) or (7 and 8), there are 5 options on how to order the remaining chairs, making for 10 ways (the white chair can't be directly next to the white chairs).
For the remaining 5 spots, there are only 4 ways to put the remaining white chair, making for 20 ways.
This means that there are \(5 \times 4 + 5 + 5 = 30\) ways if only 2 white chairs are next to each other.
Next, there are only 6 ways for all 3 white chairs to be next to each other (1,2, and 3; 2, 3, and 4; and so on).
This makes for \(56 - 30 - 6 = \color{brown}\boxed{20}\) ways.
