A gardener is planning a garden layout. There are two rectangular beds

Question

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.

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.

Sajjad1994 · Answer

Official Explanation

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.

sw999826 · Answer

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!

Apt0810 · Answer

Bed A: 7C1*6C2*4C2 = 630

Bed B: 7C2(6C1*4C2+6C2*4C1+6C3)=2436

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Sajjad1994 · Answer

Hello  1219bob1219

The complete topic is shadowed here.

Thank you!

Sajjad1994 · Answer

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 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.

GM_GMAT · Answer

For Bed B, why does 7C2 * 6C1 * 9C2 not work? 9C2 because after choosing one annual 4 Perennial and 5 Annual left to be chosen from. Thank you.

mariajosemasd · Answer

easy way, divide the highest numbers by 21. it s tough to do all that maths calculation in 2 minutes.

Shadyshades · Answer

I was confused whether I can use the same flower, which I have already put in Bed A ,to Bed B. 
Not sure if any other faces the same issue.

wasario · Answer

I unfortunately missed bed B because of silly reading mistake. However, being that this is a 800 level math type question, I will propose my explanation:

First, we know there are 7 types of shrubs, 6 types of annual flowers, and 4 types of perennial flowers, and that every bed needs to have exactly 5 flowers. We'll need this info to create the combinations formulas.

Bed A) We need exactly 1 shrub and 2 annual flowers. This means that the other 2 must be perennial flowers. We can use the combinations formula to figure out how many combos we can make of 1, 2, and 2. Taking this we get:

7C1 * 6C2 * 4C2 = 630

Bed B) This is trickier because the requirement is exactly 2 shrubs and AT LEAST 1 annual flower. This means that we have 3 cases to consider:

Case 1: Only 1 annual flower. Our bed will be 2 shrubs, 1 annual flower, and 2 perennial flowers. 
 Case 2: 2 annual flowers. Our bed will be 2 shrubs, 2 annual flowers, and 1 perennial flower. 
 Case 3: 3 annual flowers. Our bed will be 2 shrubs, 3 annual flowers, and no perennial flowers.

We need to calculate each of these individually and then add the results. Using combinations again, we get:

7C2*6C1*4C2 + 7C2*6C2*4C1 + 7C2*6C3*4C0 = 2436

Hence our answers are  B, F

Bismuth83 · Answer

You can use the same flower in type in both beds - this is not restricted. However, you cannot put a certain type more than once in a bed.

Hope that helped!

Bismuth83 · Answer

Let me help you with that.

It is hard to see, but you repeat some cases. Let's look at an easier problem. We have two sets of numbers A = { a_1 ,  a_2 ,  a_3 } and B = { b_1 ,  b_2 ,  b_3 }. We have to pick at least one from A and in total pick 3 different numbers.

Your logic would say that the answer is 3C1 * 5C2. However, we count things like  a_1a_2a_3  many times. This can be seen by let we pick  a_1  first. Then, we pick 2 from the rest - one of these cases is  a_2  and  a_3 . We can also pick  a_2  first and there will be a case with  a_1  and  a_3 . However, both of these cases are the same.

Hope that helped!

Anonymous · Answer

Hello everyone! Initially, I solved this incorrectly and only got the first part right. So, I tried to break it down to understand it better. I would appreciate your feedback on whether my reasoning is sound and how I can shorten this method to fit within the appropriate time frame of 2 minutes. bb chetan2u Bunuel KarishmaB Thank you in advance!

KarishmaB · Answer

The first part is straight forward as shown by solutions above.

The second part is an "at least type" of question. It might be easier to go the "All - Unfavorable" way though I doubt it saves much time here.

Bed B must contain exactly 2 types of shrub and at least 1 type of annual flower.

So 7C2 for 2 types of shrubs out of 7. Of the other 3 selections, at least one should be an annual flower. 
The number of ways in which none of the 3 is an annual flower will be 4C3 (selecting all 3 from perennial flowers)
The total number of ways of selecting the 3 is 10C3 (from all annual and perennial flowers).

So acceptable cases = 7C2 * (10C3 - 4C3) = 7*6/2 (10*9*8/6 - 4) = 21 * 116
This is a big number that will end in a 6 so it must be 2436.

HarshavardhanR · Answer

AF: 6
PF: 4
Sh: 7

Given: each bed contains exactly 5 flowers/shrubs and given that n o flower or shrub will used more than once in each bed

Bed A must contain exactly 1 type of shrub and exactly 2 types of annual flower. 
 
In this case, the remaining 2 (5 - 3) must be PFs.

Number of ways to select 1 shrub > 7C1
Number of ways to select 2 annual flowers -> 6C2
Number of ways of select 2 perennial flowers -> 4C2

Total number of combinations possible = 7C1 x 6C2 x 4C2  = 630 .

Bed B must contain exactly 2 types of shrub and at least 1 type of annual flower.

Here, there are 3 possibilities ->

(1) 2S, 3AF
(2) 2S, 2AF, 1PF
(3) 2S, 1AF, 2PF

Number of ways possible

(1) 2S, 3AF -> 7C2 x 6C3
(2) 2S, 2AF, 1PF -> 7C2 x 6C2 x 4C1
(3) 2S, 1AF, 2PF -> 7C2 x 6C1 x 4C2

Total number of combinations possible = (1) + (2) + (3) =  2436 .

Harsha

A gardener is planning a garden layout. There are two rectangular beds

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Bed A	Bed B
		520
		630
		756
		1,028
		2,242
		2,436

GMAT Two-part Analysis (TPA) Questions

A gardener is planning a garden layout. There are two rectangular beds

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