A bar is creating a new signature drink. There are five possible alcoh

Take the task of making a drink and break it into stages.

Stage 1: Select two alcoholic ingredients for the drink
Since the order in which we select the ingredients does not matter, we can use combinations.
We can select 2 ingredients from 5 ingredients in 5C2 ways (10 ways)
So, we can complete stage 1 in 10 ways

If anyone is interested, we have a video on calculating combinations (like 5C2) in your head (see below)

Stage 2: Select two non-alcoholic ingredients for the drink
We'll use combinations again.
We can select 2 ingredients from 5 ingredients in 5C2 ways (10 ways)
So, we can complete stage 2 in 10 ways

By the Fundamental Counting Principle (FCP), we can complete the 2 stages (and thus create a drink) in (10)(10) ways (100 ways)

Answer:

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT, so be sure to learn it

RELATED VIDEOS

5 C2 * 5 C2 = 100

Option D

Can you explain why the solution isnt 5!/2!3! ?

Since the questions asks for --> If the bar uses two alcoholic ingredients and two non-alcoholic ingredients

You would need to consider 2 selections

A bar is creating a new signature drink. There are five possible alcoholic ingredients in the drink: rum, vodka, gin, peach schnapps, or whiskey. There are five possible non-alcoholic ingredients: cranberry juice, orange juice, pineapple juice, lime juice, or lemon juice. If the bar uses two alcoholic ingredients and two non-alcoholic ingredients, how many different drinks are possible?

The question ask for the combination of 2 alcoholic ingredient selection and 2 non-alcoholic ingredient selection. There is a "and" in-between !!
and means multiply.

Now calculate the selection of 2 alcoholic ingredient first. It is given by - 5!/2!*3!.
Similarly, calculate the selection of 2 alcoholic ingredient. It is given by - 5!/2!*3!.

now we need to multiply both to get the final answer, which is given by = selection of 2 alcoholic ingredient* selection of 2 non-alcoholic ingredient i.e. (5!/2!*3!)*(5!/2!*3!).

Hope this helps.

1. There are four slot : 2 for alcoholic and 2 for non alcoholic.

2. For alcoholic, we can count for 2 slots : \(\frac{(5 X 4)}{(2 X 1)}\) = 10
We use 2 X 1 in denominator because we should eliminate the 2 double counting (Rum and Vodka combination is same with Vodka and Rum combination)

3. For non alcoholic, we can count in the exactly same way with non alcoholic. Total possibilites : 10.

4. Combine (2) and (3), we get 10 X 10 = 100. Therefore D.

Thanks.
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The number of ways we can choose two alcoholic ingredients from five is 5C2 = 5!/[2!(5-2)!] = (5 x 4)/2! = 10, and the number of ways we can choose two non-alcoholic ingredients from five is the same: 5C2 = 10. Thus, the total number of ways we can choose two alcoholic ingredients and two non-alcoholic ingredients is 10 x 10 = 100.

Answer: D
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No of possible combinations = 5C2 (ways of choosing 2 alcoholic drinks from 5 alcoholic drinks) x 5C2(ways of choosing 2 non-alcoholic drinks from 5 no-alcoholic drinks)= 10 x 10 = 100.

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