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!
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.
630 (Bed A)
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:
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.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 Ã— 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 Ã— 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 Ã— 21 = 2,436 possibilities for bed B.
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.