first draw the hexagon

Place any where a and b places

a-link-b 1 when link 0;

a-link-b 4 when link 1;

a-link-b 12 when link 2;

a-link-b 24 when link 3;

a-link-b 24 when link 4;

link is the no.of cities between a and b

So total ways from a to b are 65.

Ways No. of Ways

A B 1

A _ B The blank can be filled in 4 ways 4

A _ _ B The first blank can be filled in 4 ways

The second blank can be filled in 3 ways 4*3 12

A _ _ _ B The first blank can be filled in 4 ways

The second blank can be filled in 3 ways

The third blank can be filled in 2 ways 4*3*2 24

A _ _ _ _ B The first blank can be filled in 4 ways

The second blank can be filled in 3 ways

The third blank can be filled in 2 ways

The fourth blank can be filled in 1 way 4*3*2*1 24

So total no. of ways 1+4+12+24+24 65 ways to travel from A to B

somebody please give a better explanation

There are totally 6 cities and we have to go to City B from City A. This can be achieved by the following ways.

1. We can go directly from A to B without any city in between. So there is only one Way.

2. We can go through any one of the remaining 4 cities. Say remaining four cities be C D E & F. ie A-C-B or A-D-B or A-E-B or A-F-B. Totally 4 ways ( 4P1 ways)

3. We can go through any two of the remaining 4 cities. Like A-C-D-B or A-D-C-B or A-C-E-B and so on.. That 2 permutation out of 4 (4P2 ways). Totally we will get 4 x 3 12 ways.

4. We can go through any three of the remaining 4 cities. 3 permutations out of 4 (4P3 ways ) 4 x3 x 2 24 ways.

5. We can go through all the remaining 4 cities. Simillarlly that can be arranged in 4C4 ways 24 ways.

Hence the total no of ways 1 + 4 + 12 + 24 + 24 65 ways.

Kindly comment whether understood.