I have a very hard maths problem that I have no idea on where to start. It seems easy, but I'm sure it isn't.

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 have a very hard maths problem that I have no idea on where to start. It seems easy, but I'm sure it isn't.
Since you have to use at least three of each kind to make a bunch of 10, that means you will have to use 4 of one kind and 3 of each of the other two. So you can count those possibiliities:





4 banksias:


6C4 * 5C3 * 4C3 = 15*10*4 = 600


4 wattles:


6C3 * 5C4 * 4C3 = 20*5*4 = 400


4 waratahs:


6C3 * 5C3 * 4C4 = 20 * 10 * 1 = 200





Then add them up to get 1200 possible bunches!
Reply:There's probably a formula for this, but not one I remember.





So to start, set aside the known portion -- every bunch will have 3 banksias, 3 wattles, and 3 waratahs. That gives you 9 known flowers in each bunch.





What's remaining? 3 banksias, 2 wattles, and 1 waratah. Since you only can use 1 more flower and all the flowers of each kind are different, the possible number of combinations is the number of remaining flowers = 6
Reply:She can have:


3 banksias, 3 wattles, and 4 waratahs


3 banksias, 3 waratahs, and 4 wattles


3 waratahs, 3 wattles, and 4 banksias
Reply:Out of the 6 banksias chooose 3. That is 6C3=20.


Out of the 5 wattles choose 3. That is 5C3=10.


Out of the 4 waratahs choose 3. That is 4C3=4.





You now have 3*3 flowers, and you are guaranteed to have at least 3 of each kind.





You need one more flower, where do you get that from the bunch of those not chosen, that is 3banksias, 2wattles %26amp; 1waratahs. Thus out of 6 flowers choose 1. Then 6C1=6.





Just multiply all of them (since you need 10).


ANSWER=6C3*5C3*4C3*6C1


=20*10*4*6=4800 ways.
Reply:she has to take at least three of all types and then she was left with 3,2,1 flower of each type and according to your question each flower is different so last flower is chosen from 6 different flowers so total numbers oof ways are





=(6C3)(5C3)(4C3)(6C1)


=4800
Reply:This is a field of math called "combinatorics".





When you pick x flowers out of a total group of n flowers, you have n ways to pick the first flower but n-1 flowers remain for the 2nd pick, etc.





The number of permutations (ways of choosing) = P(n,x)





= n! / x!





where n! means "factorial" and means n multiplied with all the integers less than n, down to 1:


n! = n*(n-1)*(n-2)...*3*2*1.





So if you have 6 different banksias and pick 3, the ways to do it are


6!/3! = 6*5*4*3*2*1 / 3*2*1 = 6*5*4 = 120.





But some of the ways you picked give the same 3 flowers as other ways. You just pick the same flowers in a different order. With 3 flowers there are 6 different ways to put them in order. When you don't care what order you pick the flowers in, just the resulting flowers you get, the number of ways of choosing are called "combinations".





C(n,x) = n! / [ x! (n-x)! ]


In this case, C(6,3) = 6! / [ 3! (6-3)!]





A scientific calculator will have an n! key and maybe even C(n,x) and P(n,x) keys.





but 6! / 3! is just 6*5*4 again, and (6-3)! = 3! = 6.


C(6,3) = 20.














You are told you must pick 10 and only 10 flowers, but there must be at least 3 of each kind.





WAYS TO DO THAT:





3 out of 6 banksias, 3 out of 5 wattles and 4 out of 4 waratahs





3 out of 6 banksias, 4 out of 5 wattles and 3 out of 4 waratahs





4 out of 6 banksias, 3 out of 5 wattles and 3 out of 4 waratahs





EXPRESSED IN COMBINATIONS:





C(6,3) * C(5,3) * C(4,4) = 20 * 10 * 1 = 200.


C(6,3) * C(5,4) * C(4,3) = 20 * 5 * 4 = 400.


C(6,4) * C(5,3) * C(4,3) = 15 * 10 * 4 = 600.





Sum the combinations from these 3 possible cases and you get 1,200 ways to arrange the flowers.





I got a "thumbs-down" but I was editing my answer and corrected mistakes as I went along, so I didn't really deserve that!
Reply:She has 17 ways as follows:


Each number in a set is arranged in the order of banksias, wattles and waratahs, respectively.


a) 6+3+1


b) 6+2+2


c) 6+1+3


d) 5+4+1


e) 5+3+2


f) 5+2+3


g) 5+1+4


h) 4+5+1


i) 4+4+2


j) 4+3+3


k) 4+2+4


l) 3+5+2


m) 3+4+3


n) 3+5+2


o) 2+5+3


p) 2+4+4


q) 1+5+4


In each way, the remainder will be five flowers and will not be enough to make another bunch with a minimum of 10 per bunch.
Reply:because of this "all the flowers of each are different"


you must come up with all posibilities asuming that


every flower is of different type even though they have


the same name with in their group.





Out of the 6 banksias chooose 3. That is 6C3=20 posibilities


Out of the 5 wattles choose 3. That is 5C3=10 posibilities


Out of the 4 waratahs choose 3. That is 4C3=4 posibilities





You now have 9 flowers. You need one


more flower, which you will get from those


not taken 3 banksias, 2 wattles %26amp;


1 waratahs. So now out of 6 flowers


choose 1. 6C1=6 posibilities





Now just multiply all of them since you


need 10 flowers in each arrangement.


6C3*5C3*4C3*6C1=?


20*10*4*6= 4800


4800 different arrangements
Reply:After she uses that many : she'll have 3 bnakias 2 watteles and 1 waraths.

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