combinations

[carebear]

A lady gives a dinner party for 6 of her 9 friends. How many ways can she choose her 6 guests? I think it is C(9, 6) = 84 ways

The second part has me "stumped"....How many ways can she do this if Dorothy and Lori cannot attend together?

Where do I start? Is it like permutations.....would I somehow do "total" minus "together"? But how do I express that Dorothy and Lori attend together?

Please help.

 

[DrSteve]

There are C(7.6) = 7 ways she can choose that exclude both Dorothy and Lori.
There are C(7,5) = 21 ways she can choose that include Dorothy (she chose Dorothy, now she's gotta choose 5 more out of 7 because Lori must be exluded)
The last computation also gives the number of choices that include Lori.

So altogether we get C(7,6) + 2C(7,5) = 7 + 2*21 = 7 + 42 = 49.

Another way:

There are C(7,4) = 35 ways that Dorothy and Lori DO attend together (Once we insist that Dorothy and Lori attend we need to choose 4 more out of the remaining 7)

So subtract 84 - 35 = 49.