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Permutaion or combination

Expert replies
by sanjib » Tue May 26, 2009 2:28 pm
Katent needs to plant a group of four different plants each for her three flowerbeds. If she has twelve different plants ,how many different arrangements of plant could she have in her garden?


Is it a Permutation problem or Combination. because it asks arrangements in the last line- which means permutation

But answer is 34650
as 12C4.8C4.4C4
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Source: — Problem Solving |

Re: Permutaion or combination

by Stuart@KaplanGMAT » Tue May 26, 2009 2:44 pm
sanjib wrote:Katent needs to plant a group of four different plants each for her three flowerbeds. If she has twelve different plants ,how many different arrangements of plant could she have in her garden?


Is it a Permutation problem or Combination. because it asks arrangements in the last line- which means permutation

But answer is 34650
as 12C4.8C4.4C4
It's not a perfectly worded question, but it's a bit of both.

When we look at the entire garden, order DOES matter. If we call the 12 plants A, B, C, D, ... L, putting ABCD in 1, EFGH in 2 and IJKL in 3 is a different arrangement than ABCD in 1, IJKL in 2 and EFGH in 3.

When we look at each individual flower bed, however, order does NOT matter. We're just selecting which plants go in which bed, not the order of how we plant them in that bed.

In fact, we know that order inside the bed can't possibly matter, because we don't have enough information to determine how many different ways the 4 flowers could be planted. Are we planting them in a straight line? In a circle? In a star? Different physical arrangements would lead to different numbers of permutations.

So, the solution provided is correct:

Bed 1: 12 total plants, selecting 4 of them = 12C4
Bed 2: 8 plants left, selecting 4 of them = 8C4
Bed 3: 4 plants left, selecting 4 of them = 4C4

When we're making MULTIPLE selections, we always MULTIPLY, so the final answer is 12C4*8C4*4C4.

Note that if we didn't care about which plants went in which bed, we would then divide this answer by 3!.

For example, if the question had been "12 people are being divided up into 3 groups of 4. How many different ways can the 12 people be so divided?", the answer would be:

12C4*8C4*4C4/3! to account for the duplications.
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