Bar Drinks

harsh.champ 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, limejuice, or lemon juice. If the bar uses two alcoholic ingredients and two non-alcoholic ingredients, how many different drinks are possible?

(A)100
(B)25
(C)50
(D)75
(E)3600

The OA is A.

Anybody,up for a drink?? :)

first calculate the number of ways of selecting two alcoholic and two non-alcoholic ingredients. order of arrangement does not matter.

The number of combinations of n objects taken r at a time is
C(n,r) = n!/(r!(n-r!))
The number of combinations of alcoholic ingredients is
C(5,2) = 5!/(2!(3!))
C(5,2) = 120/(2(6))
C(5,2) = 10
The number of combinations of non-alcoholic ingredients is
C(5,2) = 5!/(2!(3!))
C(5,2) = 120/(2(6))
C(5,2) = 10
The number of possible drinks is
= 10 * 10
= 100
Ans A

harsh.champ 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, limejuice, or lemon juice. If the bar uses two alcoholic ingredients and two non-alcoholic ingredients, how many different drinks are possible?

(A)100
(B)25
(C)50
(D)75
(E)3600

The OA is A.

Anybody,up for a drink?? :)

Choose 2 out 5 * choose 2 out of 5
5C2*5C2 = 100

harsh.champ 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, limejuice, or lemon juice. If the bar uses two alcoholic ingredients and two non-alcoholic ingredients, how many different drinks are possible?

(A)100
(B)25
(C)50
(D)75
(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: A