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

Hope this helps.
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Bunuel
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.

Thanks.
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Bunuel
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.

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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