Counting #2

This has a really nice method. 

B will be left of A 50% of the time, right of A 50% of the time. 

5*4*3*2*1 total ways to be in a line, 50% of which are where B is on the left of A. 

5*4*3*2*1/2 = 60

=^._.^=

oops lemme recalculate

sure!

If B is adjacent to A we can make them as 1 letter, so  4!=24

If B is one away from A, we have 4 cases.

BCAD

BDAC

DBCA

CBDA

If B is two away from A we get:

BCDA

BDCA

24+4+2=30

30 ways

there are 5 numbers not 4

Lets have a look at your method mathisopandcool

XXXXB       4!             = 24

XXXBX       3*3!          =18

XXBXX       3*2! *2      =12

XBXXX      3!              =6

total = 24+18+12+6 = 60 ways

If you want me to explain any of these just ask.    

A  X  X  X  X       Four ways for B to be to the right

X A X X X           Three ways

X X A X X            Two ways

X X X A X              One way     total of 10 ways     

                                      the other three slots can be filled in   3 * 2 *1   ways  = 6 ways for EACH of the 10 possibles

      10 * 6 = 60 ways

ok ok thanks for all the help!