Combinatorics

[mariar827]

Hi, I was checking the number properties Manhattan Prep Guide and found this problem that is making my life impossible:
a florist has 2 azaleas, 3 buttercups and 4 petunias. She puts two random flowers in a bouquet. However the costumer calls and says that she does not want two of the same flower. What is the probability that the florist won't have to change the bouquet?
Actually, i understand how to solve it, i have to find the probability of a bouquet with two of the same flowers, and then 1-that value, will be the probability I'm looking for.
I can't find the number of two flower bouquets to be made, as my reasoning leads me to this: 9!/(2!3!4!) and i know it incorrect, using the formula for combinations I arrive to the answer, but i don't understand the reasoning of the book which is something like:
Then, find the number of
different 2-flower bouquets that can be made in total, using an anagram
model. In how many different ways can you arrange the letters in the
“word” YYNNNNNNN? which leads to 9!/(7!2!)
i mean if i have 3 types of flower, why the anagram only considers two?, shouldn't it be like YYBBBNNNN
I find easier to solve my problems with the MGMAT technique rather than using the formula.
Thanks a lot for the help!!!

[RonPurewal]

hi,
please re-post this query in the correct folder (= the MGMAT non-CAT math folder).

thanks.