A florist has to make a floral arrangement. She has 6 banksias, 5 wattles and 4 waratahs. All the flowers of each kind are different. In how many ways can the florist make a bunch of 10 flowers if she has to use at least 3 of each kind?
I know it has been asked before but the answer given was unclear. Any help appreciated thanks
Math Combinations/Permutations Question?
First 3 positiion = Banksias
So 6 x 5 x 4 = 120
Second 3 Positions = Wattles
5 x 4 x 3 = 60
Third 3 Positions = Waratahs
So 4 x 3 x 2 = 24
120 x 60 x 24 = 172800
There must be 3 banksias, 2 wattles and 1 waratah left to take last position
therefore 6 ways to take last space
So 172800 x 6
= 1036800
Reply:3*3*3*3*3*3*3*3*3*3 = Answer
Reply:for a bunch of 10 flowers ( at least 3 from each kind of A ( 6 ), B ( 5 ). C ( 4 ) ways are = C(4,4)*C(5,3)*C(6,3)
+ C(4,3)*C(5,4)*C(6,3)
+ C(4,3)*C(5,3)*C(6,4)
= 1*10*20 + 4*5*20 + 4*10*15
= 200 + 400 + 600
= 1200 ways
Reply:permutations are an ordered set without repititions. combinations are an unordered set of unique combinations. For example, both 2,3,5 and 3,2,5 are acceptable subsets of a permutation. However, a combination considers these to be the same subset.
In your case, with the flowers, either the question is very simple and the answer is three, or it%26#039;s very complex and I don%26#039;t feel like getting into it.
Reply:Basically you need to determine how many combinations per type of flower there are. Then you know you can %26quot;combine%26quot; those seperate %26quot;choices%26quot;/combinations together to get your 10 flowers.
So (6 choose 3) = # of ways to arrange those 6 flowers (# of ways to combine them/choose)
Do the same for the other two groups of flowers (5 choose 3) and (4 choose 3).
Finally you need to select 1 flower. We have already choose 9 of the total 15, so 6 are left. So the last flower would be (6 choose 1) which is 6.
So the total number of ways to choose 10 flowers would be:
(6 choose 3) * (5 choose 3) * (4 choose 3) * 6
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