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# How many different combinations can be made from 5 varieties of flowers?

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If I have 5 different varieties of flowers:
carnations, roses, mums, marigolds and lilies.
And I put 3 different varieties of flowers in a vase
For example, a vase might contain roses, carnations and lilies
How many different combinations can be made from the 5 varieties of flowers?

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Commented:
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Commented:
You can use combinatory analysis

Cx,y =   x!
_________
y! * (x - y)!

In you example:

C5,3 =        5!              =   120      =   120    = 120   = 10
__________      _______     _____    ____
3! * (5 - 3)!          6 * 2!        6 * 2       12
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Commented:
You want to pick 3 kinds of flowers out of 5. So that is:
5 things taken 3 at a time = 5 C 3 = 5!/[(5-3)!(3!)] = 5*4/2! = 20/2 = 10 combinations.
http://www.mathwords.com/c/combination_formula.htm
http://www.wolframalpha.com/input/?i=5+C+3

Commented:
Does this equate to:

5!                                     120
--                                      ---        =  10
3! * (5 - 3)!          =            6 * 2
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Commented:
Yes, same calculations as above.

Note that 5! = 5*4*3!
so that the 3! in the denominator cancels out with the 3! in the numerator, and then you have

5!
------------  = 5*4/2! = 20/2 = 10
3! * (5 - 3)!
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Commented:
So, in general, taking R things out of N leads to this number of combinations:

N!             N(N-1)...(N-R+1)(N-R)!
----------- = ------------------------------- = N(N-1)...(N-R+1)/R!
R!( N-R )!                          R!( N-R )!

Here the (N-R)! factor in the numerator cancels with the same factor in the denominator.
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Commented:
Here is the general form that I used with the concrete numbers:

N!             N(N-1)...(R+1)(R)!
----------- = -------------------------- = N(N-1)...(R+1)/( N-R )!
R!( N-R )!                     R!( N-R )!

Here the R! factor in the numerator cancels with the same factor in the denominator.
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