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A bar is creating a new signature drink. There are five possible alcoh
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14 Apr 2017, 03:40
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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?
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14 Apr 2017, 06:44
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Bunuel wrote:
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?
A. 25 B. 50 C. 75 D. 100 E. 3600
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)
A bar is creating a new signature drink. There are five possible alcoh
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18 Apr 2017, 00:44
chinelj wrote:
Can you explain why the solution isnt 5!/2!3! ?
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!).
A bar is creating a new signature drink. There are five possible alcoh
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24 Jul 2017, 18:54
Bunuel wrote:
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?
A. 25 B. 50 C. 75 D. 100 E. 3600
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.
Re: A bar is creating a new signature drink. There are five possible alcoh
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26 Jul 2017, 16:36
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Bunuel wrote:
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?
A. 25 B. 50 C. 75 D. 100 E. 3600
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.
Re: A bar is creating a new signature drink. There are five possible alcoh
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18 Dec 2018, 23:33
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.
gmatclubot
Re: A bar is creating a new signature drink. There are five possible alcoh
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18 Dec 2018, 23:33