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Mary has 6 identical basil plants, and three different window sills she can put them on. How many ways are there for Mary to put the plants on the window sills?

 Dec 6, 2017
 #1
avatar+128079 
+1

This is analogous to filling  N distinct boxes with K identical balls

 

If  there are no restrictions as to the number of plants each sill can contain ....the number of possible  ways   is

C( K + N - 1, N - 1)  = 

C ( 6 + 3 - 1, 3 - 1)  =

C (8, 2) = 28 ways

 

If each sill must contain at least one plant, the number of possible ways is

C( K - 1, N - 1)  =

C( 6 - 1, 3 - 1)  =

C(5, 2)  =  10 ways

 

 

cool cool cool

 Dec 6, 2017
edited by CPhill  Dec 6, 2017
 #2
avatar+488 
+6

By the way that is incorrect

Real Solution

 

Since the plants are indistinguishable, we must only count the number of plants on each window sill.

If 5 all the plants are on one window sill, there are  ways to choose which window sill they are on.

If 5 plants are on one window sill and the last is on another, there are  ways to choose which plants go on which window sill.

If 4 plants are on one window sill and the last two are on another, there are  ways to choose which window sill they are on.

If 4 plants are on one window sill and the last two are each on one of the other windows, there are  ways to choose which window the plants are on.

If 3 plants are on one window and the other  plants are all on another window, there are  ways to choose which window has no plants.

If 3 plants are on one window,  plants on another window, and  plant on the last window, there are  ways to choose which plants are on which windows.

If 2 plants are on each window, there is only one way to arrange them.

In total, there are \(3+6+6+3+3+6+1=\boxed{28}\) ways to arrange the plants on the window sills.

 

RektTheNoob  Dec 9, 2017
edited by RektTheNoob  Dec 9, 2017
 #3
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-5

If you are so certain of the answer, why did you pose it in the first place ????!!!!

Guest Dec 9, 2017
edited by Guest  Dec 9, 2017

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