Counting Review (Chapter 4) Math2320

Consider a set of 6 people to be seated. (a) How many ways can they be seated on a bench with six seats? 'People' are assumed distinct! We can either (i) analyse where each of the people sits as they walk up or (ii) the way the seats 'choose' who sits on them! (The answers should be the same.)
(i)            The first person in has  6 places to sit;
    and   the second person in has  5  remaining places to sit;
    and   the third person in has  4  remaining places to sit;
    and   the next person in has  3  remaining places to sit;
    and   the next person in has  2  remaining places to sit;
    and   the last person in has  1  remaining place to sit.

Since  and  becomes multiplication, we have  
       6x5x4x3x2x1 = 6!  =  P(6,6)  ways to seat the people.

(ii)    The first seat can be filed  by a person in  6 ways;
    and   the second seat can be filed by   5  remaining people;
    and   the third seat can be filed by   4  remaining people;
    and   the next seat can be filed by   3  remaining people;
    and   the next seat can be filed by   2  remaining people;
    and   the next seat can be filed by   1  remaining person.

Since  and  becomes multiplication, we have  
       6x5x4x3x2x1 = 6!  =  P(6,6)  ways to seat the people.

(b) How many ways can they be seated around a rectangular table with one seat on each end and two along each side? With the table, we realize that two 'fixed seatings' will be 'the same', since we can rotate the entire table by 180 degrees without making any difference! Thus we can count the seatings ignoring this symmetry and then apply the 'equivalence principle': two of these seatings are 'equivalent' so our initial answer should be divided by 2! We can again either (i) analyse where each of the people sits as they walk up or (ii) the way the seats 'choose' who sits on them! (The answers should be the same.)

(i)            The first person in has  6 places to sit;
    and   the second person in has  5  remaining places to sit;
    and   the third person in has  4  remaining places to sit;
    and   the next person in has  3  remaining places to sit;
    and   the next person in has  2  remaining places to sit;
    and   the last person in has  1  remaining place to sit.

Since  and  becomes multiplication,
and the equivalence principle says to divide by 2,  we have  
       6x5x4x3x2x1/2 = 6!/2!  =  3x5x4x3x2x1   ways to seat the people.

Alternatively, we can wash out the equivalence by realizing that the
first person in really has only 3 places to sit:  at an end, at the right of an end
or to the left of an end!  The rest of the people now have the same choices 
as before (with someone seated  all  remaining seats are
clearly different).
  The first person in has  3 places to sit;
    and   the second person in has  5  remaining places to sit;
    .   .   .   .  
    and   the last person in has  1  remaining place to sit.

There are  3x5x4x3x2x1   ways to seat the people.

(ii) This is   harder  to do by  having the seats 'choose' people, since
there are two seats which start off the same!  With this approach we will
have to use the equivalence principle!