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IR Question from MGMAT IR Bank

This topic has 1 expert reply and 1 member reply
vrchen Junior | Next Rank: 30 Posts Default Avatar
Joined
29 Jul 2012
Posted:
11 messages
Test Date:
9/6/2012

IR Question from MGMAT IR Bank

Post Sat Aug 25, 2012 11:23 pm
Hi All,

I came across an combinatorics IR question in the MGMAT IR question bank and was wondering why my method doesn't work. I've pasted the question + answer explanation along with my method below. It'd be great if someone could help me understand this better. Thanks in advance!

Question:
Quote:
A gardener is planning a garden layout. There are two rectangular beds, A and B, that will each contain a total of 5 types of shrubs or flowers. For each bed, the gardener can choose from among 6 types of annual flowers, 4 types of perennial flowers, and 7 types of shrubs. Bed A must contain exactly 1 type of shrub and exactly 2 types of annual flower. Bed B must contain exactly 2 types of shrub and at least 1 type of annual flower. No flower or shrub will used more than once in each bed.
Identify the number of possible combinations of shrubs and flowers for bed A and the number of possible combinations of shrubs and flowers for bed B.

Make only two selections, one in each column.
520
630 (Bed A)
756
1,028
2,242
2,436 (Bed B)
I understood how to derive the answer for Bed A but could not for Bed B.

Here's MGMAT's explanation for Bed B:


Quote:
For Bed B, we’re told that the gardener must include exactly 2 types of shrub and at least 1 type of annual flower. This bed must also contain a total of 5 different types of shrubs or flowers.

To choose 2 shrubs from 7 possibilities, we calculate 7!/(5!2!) = 21.
Choosing the bed B flowers is more tricky. We have three possible scenarios: the gardener chooses 1 annual flower (and therefore 2 perennials) OR the gardener chooses 2 annuals (and therefore 1 perennial) OR the gardener chooses 3 annuals (and therefore 0 perennials). We need to calculate the number of combinations for each and then add them together.

1 annual and 2 perennials: choosing 1 always matches the number of options, so there are 6 ways to choose the 1 annual flower. We calculated the number of possible combinations for 2 perennials when we did bed A: the number of possible combinations is 6. There are, therefore, 6 X 6 = 36 possible ways to have 1 annual and 2 perennials.

2 annuals and 1 perennial: We calculated the number of possible combinations for 2 annuals when we did bed A above: the number of possible combinations is 15. For the perennials, we’re choosing only 1, so there are 4 possible ways to choose a perennial. There are 15 X 4 = 60 possible ways to have 2 annuals and 1 perennial.

3 annuals and 0 perennials: to have three annuals, we calculate 6!/(3!3!) = 20.

To have 1 OR 2 OR 3 annuals, we have 36 + 60 + 20 = 116 possible ways. In order to have this AND our 2 shrubs, we have 116 X 21 = 2,436 possibilities for bed B.
What I don't specifically understand is why we have to break out the 3 calculations for the combination of annual and perennial flowers. The way I went about doing this is half slot method and half "choosing a team" method.

I thought of Bed B has having 5 slots to fill with different types of flowers. Since we know we have to have exactly 2 shrubs, I allocated to slots to "S":

5 slots: _ _ _ _ _

allocate 2 to schrubs: S S _ _ _

Since the problem said there were has to be at least 1 annual flower, I allocated one slot for "A" then left the remaining 2 spots open for either annual or perennial flowers:

S S A A/P A/P

Now I think about each type of slot and how many combinations is possible:

2 Slots for S, can select from 7 types = 7!/(2!5!) = 21
1 Slot for A, can select from 6 types = 6!/(1!6!) = 6
2 Slots for A or P, can select from 9 types (5 remaining A's + 4 P's) = 9!/(2!7!) = 36

Then multiply it all out: 21x6x36 = 4536 combinations (WRONG)

But this answer is obviously not consistent with the solution.

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Jim@StratusPrep MBA Admissions Consultant
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Post Sun Aug 26, 2012 8:15 am
You are not eliminating duplicates in your solution. For example you include both the the following groupings even though they are the same:

a1 a2 p1

a2 a1 p1

This is why you have to break it up into the smaller groups as per the explanation.

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Thanked by: vrchen
vrchen Junior | Next Rank: 30 Posts Default Avatar
Joined
29 Jul 2012
Posted:
11 messages
Test Date:
9/6/2012
Post Mon Aug 27, 2012 9:11 pm
Thanks Jim!

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