(i) The first person in has 6 places to sit;andthe second person in has 5 remaining places to sit;andthe third person in has 4 remaining places to sit;andthe next person in has 3 remaining places to sit;andthe next person in has 2 remaining places to sit;andthe last person in has 1 remaining place to sit. Sinceandbecomes 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;andthe second seat can be filed by 5 remaining people;andthe third seat can be filed by 4 remaining people;andthe next seat can be filed by 3 remaining people;andthe next seat can be filed by 2 remaining people;andthe next seat can be filed by 1 remaining person. Sinceandbecomes 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;andthe second person in has 5 remaining places to sit;andthe third person in has 4 remaining places to sit;andthe next person in has 3 remaining places to sit;andthe next person in has 2 remaining places to sit;andthe last person in has 1 remaining place to sit. Sinceandbecomes 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 seatedallremaining seats are clearly different). The first person in has 3 places to sit;andthe second person in has 5 remaining places to sit; . . . .andthe last person in has 1 remaining place to sit. There are 3x5x4x3x2x1 ways to seat the people. (ii) This isharderto 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!