Thread: 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

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

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