Question 366512: Mary purchased a package of 18 different plants, but she only needed 12 plants for planting. In how many ways can she select the 12 plants from the package to be planted?
Here is what I got:
18!/12!= 18*17*16*15*14*13= 13366080
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
No. But the correct answer depends on whether order is significant in differentiating between selections.
If you consider the plants as being numbered 1 through 18, and that one of the selections was 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 -- in that order and that selection was considered different than, say, 7, 8, 9, 10, 11, 12, 1, 2, 3, 4, 5, 6, which is the same selection but in a different order, then the formula you need to use is:
\ =\ \frac{n!}{(n\ -\ r)!})
Which, for your problem becomes:
\ =\ \frac{18!}{(18\ -\ 12)!}\ =\ 8,892,185,702,400)
However, if the two examples I gave (plus all other possible arrangements of the first 12) are considered the same selection, then you need another factor in your denominator:
\ =\ \frac{n!}{r!(n\ -\ r)!})
\ =\ \frac{18!}{12!(18\ -\ 12)!}\ =\ 18,564)
John
My calculator said it, I believe it, that settles it
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