permutation/combination question again

5 people stand in line..

A)how many way are there to arrange these people

B)if A and B must stand together, how many ways are there to arrange these people

C)what is the probability A and B stand together

i think i may have this one figured out, but would like some confirmation that i did it correctly

A) 5!=120

B) i grouped A and B together making my sample AB,C,D,E

-then i did P4/4=4!=24

-then i considered that AB can be grouped in two ways so did p2/2=2!=2

-then i multiplied the results 24x2=48 combinations

C) 48/120

is this correct?

i appreciate any input

thank you

-ray

Re: permutation/combination question again

you are right

an other method for (B)

AB c d e

c AB d e

c d AB e

c d e AB

we have 4 x 2 = 8 posibilities and for each (3!) => 8 x 3! = 8 x 6 = 48

Re: permutation/combination question again

Quote:

Originally Posted by

**salim** you are right

an other method for (B)

AB c d e

c AB d e

c d AB e

c d e AB

we have 4 x 2 = 8 posibilities and for each (3!) => 8 x 3! = 8 x 6 = 48

thank you