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For a fundraising dinner, a florist is asked to create

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by gmat_winter » Wed Mar 25, 2015 9:55 am
For a fundraising dinner, a florist is asked to create flower arrangements for 8 tables. Each table can have one of the two types of bouquets available, one with a single type of flower or one with three different types of flowers. If the florist wants to make each table unique, what is the least number of types of flowers he needs?
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by GMATGuruNY » Wed Mar 25, 2015 10:23 am
gmat_winter wrote:For a fundraising dinner, a florist is asked to create flower arrangements for 8 tables. Each table can have one of the two types of bouquets available, one with a single type of flower or one with three different types of flowers. If the florist wants to make each table unique, what is the least number of types of flowers he needs?
4
5
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8
Each of the 8 tables is to receive either 1 type of flower or 3 types of flower.
We can PLUG IN THE ANSWERS, which represent the least number of flower types needed.
When the correct answer choice is plugged in, at least 8 different bouquets will be possible.
Since we must determine the LEAST number of flower types needed, we should test the SMALLEST answer choice first.

Answer choice A: 4
Let the 4 types of flowers be A, B, C and D.
Bouquet options:
A
B
C
D
ABC
ABD
ACD
BCD
Total options = 8.
Success!

The correct answer is A.
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by [email protected] » Wed Mar 25, 2015 6:04 pm
Hi gmat_winter,

Mitch's approach (TESTing THE ANSWERS) is a great way to tackle this type of question. You can also solve it using the Combination Formula and doing a bit of math.

Combinations = N!/K!(N-K)! Where N is the total number of items and K is the number of items in the 'subgroup'

The question asks to form 8 different flower bouquets using groups of flowers. We can make a bouquet with just 1 type of flower or 3 different types of flowers. We're asked for the LEAST number of different types of flowers needed to create 8 different bouquets.

Let's start with the SMALLEST Answer.

With 4 different types of flower, we could have....

Bouquets with just 1 flower = 4c1 = 4!/(1!3!) = 4 different options

Bouquets with 3 different flowers = 4c3 = 4!/(3!1!) = 4 different options

Here, we have 4+4 = 8 different options, which is exactly what we're looking for. This MUST be the answer.

Final Answer: A

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