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

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.

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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