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Martha and Four Friends

```
Date: 09/17/2001 at 07:30:12
From: vidya
Subject: Combinations permutations

Martha and four friends go to a movie. How many different ways can
they sit together with Martha always between two friends?

My answer is 72, whereas the solutions gives the answer to the
question as 24.

Thanking you in anticipation,
Vidya
```

```
Date: 09/17/2001 at 09:56:06
From: Doctor Ian
Subject: Re: Combinations permutations

Hi Vidya,

I get the same answer you do. The total number of ways that they can
sit together is

5! = 5 * 4 * 3 * 2 * 1

= 120

The total number of ways that Martha can sit on the left end is

4! = 24

The total number of ways that she can sit on the right end is

4! = 24

So the total number of ways that she can sit in the middle must be

120 - 24 - 24 = 72

I'm guessing that the person who said that 24 is the answer reasoned
this way:

There are 3*2*1 = 6 ways that Martha can sit with two of her
friends on the inside. Then there are 4 ways to choose the
person on the left; which leaves just 1 way to choose the person
on the right; for a total of 4*3*2*1*1 = 24 ways for the
friends to sit together.

However, this is incorrect, since it doesn't take into account the
ways that the inner and outer groups can be chosen. But as I said, I'm
just guessing.

But in problems like this - assuming they are small enough - there is
always a way to settle a question definitively, which is to enumerate
all the possibilities. To generate them systematically, I'm going to
use the notation '-- ---', in which the left side represents choices
that have been made, and the right side represents the unchosen
elements. And let's say that Martha is 'a', and mark each permutation
in which she doesn't appear on either end.

a  bcde -> ab cde -> abc de -> abcde
abced
abd ce -> abdce
abdec
abe cd -> abecd
abedc

ac bde -> acb de -> acbde
acbed
acd be -> acdbe
acdeb
ace bd -> acebd
acedb

ae bcd -> aeb cd -> aebcd
aebdc
aec bd -> aecbd
aecdb
aed bc -> aedbc
aedcb

b  acde -> ba cde -> bac de -> bacde *
baced *
bae cd -> baecd *
baedc *

bcaed *
bcd ae -> bcdae *
bcdea
bceda

bd ace -> bda ce -> bdace *
bdaec *
bdc ae -> bdcae *
bdcea
bde ac -> bdeac *
bdeca

be acd -> bea cd -> beacd *
becda
bed ac -> bedac *
bedca

c  abde -> ca bde -> cab de -> cabde *
cabed *
cae bd -> caebd *
caedb *

cbaed *
cbd ae -> cbdae *
cbdea
cbeda

cd abe -> cda be -> cdabe *
cdaeb *
cdb ae -> cdbae *
cdbea
cde ab -> cdeab *
cdeba

ce abd -> cea bd -> ceabd *
cebda
ced ab -> cedab *
cedba

d  abce -> da bce -> dab ce -> dabce *
dabec *
dac be -> dacbe *
daceb *
dae bc -> daebc *
daecb *

db ace -> dba ce -> dbace *
dbaec *
dbc ae -> dbcae *
dbcea
dbe ac -> dbeac *
dbeca

dc abe -> dca be -> dcabe *
dcaeb *
dcb ae -> dcbae *
dcbea
dce ab -> dceab *
decba

de abc -> dea bc -> deabc *
deacb *
deb ac -> debac *
debca
dec ab -> decab *
decba

e  abcd -> ea bcd -> eab cd -> eabcd *
eabdc *
eac bd -> eacbd *
eacdb *

eb acd -> eba cd -> ebacd *
ebcda
ebd ac -> ebdac *
ebdca

ec abd -> eca bd -> ecabd *
ecbda
ecd ab -> ecdab *
ecdba

ed abc -> eda bc -> edabc *
edacb *
edb ac -> edbac *
edbca
edc ab -> edcab *
edcba

I count 72 different ways for them to sit together with Martha in the
middle.

Does this help?

- Doctor Ian, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Permutations and Combinations

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