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Joined: 06 Oct 2018
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Permutations and Combinations Question
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08 Dec 2018, 23:50
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.
Bed A Bed B Combinations 520 630 756 1,028 2,242 2,436
Here is the Official Answer:
For Bed A, we’re told that the gardener must include exactly 1 type of shrub and exactly 2 types of annual flower. The problem also specifies that bed A must contain a total of 5 different types of shrubs or flowers, so bed A must also contain 2 types of perennial flower. To choose 1 shrub from 7 possibilities, we calculate 7!/(1!6!) = 7. When choosing only 1, there is always the same number of possibilities as there is of items (7 possibilities, so 7 choices). To choose 2 annual flowers from 6 possibilities, we calculate 6!/(2!4!) = 15. To choose 2 perennial flowers from 4 possibilities, we calculate 4!/(2!2!) = 6. There are 7 possibilities for a shrub, 15 possibilities for the annual flowers, and 6 possibilities for the perennial flowers. In total, there are 7×15×6 = 630 possibilities for bed A.
My question here is why must bed A "must also contain 2 types of perennial flowers"? All it says in the question is Bed A must contain exactly 1 type of shrub and exactly 2 types of annual flower.
Here is the answer for Bed B:
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 × 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 × 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 × 21 = 2,436 possibilities for bed B.
So in Bed B they calculate the different possibilities and add them together. Shouldn't this been the method to tackle Bed A as well?



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Joined: 19 Mar 2018
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Re: Permutations and Combinations Question
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25 Dec 2018, 03:04
Hello, Thanks marathonrunner, Is there a way to answer the question more quickly please?



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Re: Permutations and Combinations Question
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02 Jan 2019, 21:22
Hello marathonrunner, Thanks for the post! 1. It says Bed A must contain 5 flowers or shrubs, and Bed A should have exactly 1 shrub and exactly 2 annuals, meaning that we have 2 places left. Since we only have 3 choices (annuals, perennials, shrubs), perennials will be our only choice (we cannot add annuals because it says we can only have 2, and we can't add shrubs because it says we can only have 1) 2. For Bed B, we are required to have 2 shrubs and AT LEAST 1 annual, again, 2 places left. But this time is different, because the number of annual can go from 1 up to 3 (we cannot add 4 annuals because the total will be more than 5), and the number of perennials change as we change the number of annuals. We need to discuss all 3 situations. If we add 1 annual (meaning that we add 2 perennials): we choose 1 from 6 types of annual: 6, and we choose 2 from 4 types of perennials: 4*3 / 2=6 (choose 1 from 4, then choose 1 from 3 left, divided by 2 because we cannot use them repeatedly. That's how I calculated, you may have your own way), times them together we get 36. For the remaining 2 situations, we do the same thing. Hope my answer help!




Re: Permutations and Combinations Question &nbs
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02 Jan 2019, 21:22






