Question

1. In how many different ways can 5 people be seated at a round table?

Answer 1

1. 

Assume the people A, B, C, D and E are sitting in a row.

since there are 5 positions, they can sit in 5*4*3*2*1=120 ways.

consider 1 certain sitting position, for example ABCDE, that is B has A to his right, C to his left. D has C to his right and E to his left. And so on.

the positions 

ABCDE
BCDEA
CDEAB
DEABC
EABCD

while in a row are different positions, in a round table they are the same thing.

(read the letters starting from A, and when the letters finish, continue reading from the beginning of the row. They all read "ABCDE")

This means that any arrangement in a round table, is converted to 5 arrangements in a row.

So there are in total  $\frac{5!}{5}=4*3*2*1=24$ arrangements of 5 people around a table.

2. check picture.

Consider the case where A is at (2, 2) and B is at (5, 6)

A can move to B through 3 horizontal units to the right, we call them E (for East) and 4 vertical units up, which we call N (for north).

the total path is 7 units.

it can be done in several ways, for example:

ENNNENE (the red path shown in the figure)
NNEEENN (the black path shown in the figure)

In total there are 
$\frac{7!}{3!*4!}= \frac{7*6*5*4!}{3!*4!}= \frac{7*6*5}{3!}= 35$ paths.

Remark: 7! is the number of arrangements if the letters were different. We divide by 3! because of the 3 E's and 4! because of the 4 N's. 

Another possibility could be A'(3, -5), B'(8,-3).

From A' to B' we go by a total of 5 E and 2 N, so there are 

$\frac{7!}{2!*5!}= \frac{7*6*5!}{2!*5!}= \frac{7*6}{2}= 42$ paths in total.
there is one more possibility
$\frac{7!}{1!(6!)}=7$

C.

a) consider the letters {V,E,C,T,O,R}

we can form 6*5*4*3*2*1=720 words, as the first digit can be any of the 6 letters, combined with 5 for the second letter and so on.

b) the difference with a is that we have 5 letters and we have 2 T.

we have in total 5*4*3*2*1=120 arrangements of these letters, if the 2 T's were considered as separate.

consider the arrangements:

$T_1RUST_2$ and $T_2RUST_1$.

We read bth as just TRUST, but the permutation formula 5*4*3*2*1 considers these as 2 different.

this is why we need to divide 120 by 2, to get the actual number of words that can be formed. So the number is 60.

Answers:

1) 24

2) 7, 35 or 42. it depends on the positions of A and B, check the solution

3) a-720, b-160