# Permutations 2

• Nov 17th 2009, 05:59 PM
john-1
Permutations 2
At a banquet, 4 coupes are sitting along one side of a table with men & women alternating.

a) how many seating arrangements are possible for these eight people?

b) how many arrangements are possible if each couple sits together? explain your reasoning.

c) how many arrangements are possible if no one is sitting beside his or her partner?

d) explain why the answers from parts b and c do not add up to the answer from part a.

a) 8!
b) 4! x 2
c) 8! - 4! x 2
d) actually, for me.. part b and c do add up. I think my work for these parts are right though..? so i don't know why they would not add up to the answers in part a. i took the total arrangements and subtracted the times that the partners are together thus i should get the times when they are not together. that's my reasoning, but i'm not sure.
• Nov 17th 2009, 06:04 PM
ANDS!
• Nov 17th 2009, 06:08 PM
john-1
for part b), I combined the couples like one unit so for eg. couple A has a man and a lady. the man and lady are now together as one unit, so count them as one instead of two since they have to always be together. so then I have 4! couples together. but each couple or unit can be either man or female or female then male. that's why there's x 2 << two possibilities within one unit.
• Nov 17th 2009, 06:16 PM
Sampras
Quote:

Originally Posted by john-1
for part b), I combined the couples like one unit so for eg. couple A has a man and a lady. the man and lady are now together as one unit, so count them as one instead of two since they have to always be together. so then I have 4! couples together. but each couple or unit can be either man or female or female then male. that's why there's x 2 << two possibilities within one unit.

Let $A = 12$, $B = 34$, $C = 56$ and $D = 78$. These are the 4 couples.

So how many arrangements are there of ABCD? I think you're right. You will probably have to use inclusion-exclusion.
• Nov 17th 2009, 06:19 PM
ANDS!
Ok thats good. So you recognize that there is a particular case where the groups are sitting next to their partners. However, is that the ONLY possible way someone can be sitting next to their partner? Think about that for a minute. You could have three couples sitting next to their partner, and one couple not; you could have two couples sitting next to their partner and two couples not. . .etc. etc. Each of those cases would have to be subtracted from the answer to part A yes?

Play with that and let me know what you get (it might take a bit of number crunching).
• Nov 17th 2009, 06:32 PM
john-1
Quote:

Originally Posted by ANDS!
Ok thats good. So you recognize that there is a particular case where the groups are sitting next to their partners. However, is that the ONLY possible way someone can be sitting next to their partner? Think about that for a minute. You could have three couples sitting next to their partner, and one couple not; you could have two couples sitting next to their partner and two couples not. . .etc. etc. Each of those cases would have to be subtracted from the answer to part A yes?

Play with that and let me know what you get (it might take a bit of number crunching).

7! x 2
+
6! x 2
+
5! x 2
+
4! x 2

= 10080 + 1440 + 240 + 48 = 11808 cases

this is the answer for part b)

part c)
if no one is sitting beside his/her partner then
8! - 11808 =28512

but now part d) isn't correct either

EDIT:
c) if no one is sitting beside his/her partner.. that means you have to take into account all four cases right? or would you just take into account the the case: 4! x 2 ? (what I had before)
• Nov 18th 2009, 07:23 AM
john-1
can ANDS! or someone else reply? I need to finish this question but am stuck on this last part. please and thank you! appreciate it
• Nov 18th 2009, 12:40 PM
john-1
does anyone know?
• Nov 18th 2009, 12:57 PM
Plato
Quote:

Originally Posted by john-1
At a banquet, 4 coupes are sitting along one side of a table with men & women alternating.

a) how many seating arrangements are possible for these eight people?

b) how many arrangements are possible if each couple sits together? explain your reasoning.

c) how many arrangements are possible if no one is sitting beside his or her partner?

d) explain why the answers from parts b and c do not add up to the answer from part a.

a) how many seating arrangements are possible for these eight people?
$2\cdot(4!)^2$

b) how many arrangements are possible if each couple sits together? explain your reasoning.
$(4!)\cdot2$

c) how many arrangements are possible if no one is sitting beside his or her partner?
$\sum\limits_{k = 0}^4 {( - 1)^k \binom{4}{k} ((4-k)!)^2\left( {2} \right)}$

Those are the answers you would find in the "back of the book".
Now you need to explain them to yourself.
• Nov 18th 2009, 03:32 PM
Soroban
Hello, john-1!

I think I got it . . .

Quote:

At a banquet, 4 coupes are sitting along one side of a table with men & women alternating.

(a) How many seating arrangements are possible for these eight people?

There are 2 cases: . $MWMWMWMW$ or $WMWMWMWM$

In each case, the 4 men can be seated in 4! ways, the 4 women can be seated in 4! ways.

There are: . $2\cdot4!\cdot4! \:=\:1152$ arrangements.

Quote:

(b) How many arrangements are possible if each couple sits together?
Call the men: . $\{A,B,C,D\}$
Let their wives be: . $\{a,b,c,d\}$, respectively.

Duct-tape the couples together.
Then we have 4 "people" to arrange: . $\boxed{Aa}\;\boxed{Bb}\;\boxed{Cc}\;\boxed{Dd}$
. . They can be arranged in $4!$ ways.

But each couple can be ordered $\boxed{Aa}$ or $\boxed{aA}$.
. . Hence, there are: . $2^4$ orderings.

There are: . $2^4 \cdot 4! \:=\:384$ arrangements.

Quote:

(c) How many arrangements are possible if no one is sitting beside his or her partner?
I found no neat formula for this . . .

The 4 men can be seated in 2 ways: . M _ M _ M _ M _ .or ._ M _ M _ M _ M

Then the 4 men can be ordered in 4! ways.

Suppose the men are seated like this: .A _ B _ C _ D

Then there are 3 seatings for the women (not adjacent to their husbands):

. . $A\; c\; B\; d\; C\; a\; D\;b \qquad A\;c\;B\;d\;C\;b\;D\;a \qquad A\;d\;B\;a\;C\;b\;D\; c$

There are: . $2 \cdot 4! \cdot 3 \;=\;144$ arrangements.

Quote:

(d) Explain why the answers from parts (b) and (c) do not add up to the answer from part (a).
Part (b) is the number of ways that all couples are adjacent.
Part (c) is the number of ways that no couples are adjacent.

These two numbers do not include the seatings where:
. . one couple is adjacent, the other 3 couples are not: (A a) B d C b D c
. . two couples are adjacent, the other 2 couples are not: (A a) B c (D d) C b
. . three couples are adjacent the other couple is not: (A a) D (b B) (c C) d