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Difficult Math Problem #61 - Combinations

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by 800guy » Wed Nov 22, 2006 6:11 pm
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
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Re: Difficult Math Problem #61 - Combinations

by kulksnikhil » Thu Nov 23, 2006 5:01 pm
800guy 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

5C2 * 5C2
= (5!/((5-2)! * 2! ) * (5!/((5-2)! * 2! )
= 10 * 10 = 100

Answer is A : 100
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by gmat_enthus » Fri Nov 24, 2006 7:27 am
Agreed. 5C2*5C2
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OA

by 800guy » Fri Nov 24, 2006 11:24 am
OA:

The first step in this problem is to calculate the number of ways of selecting two alcoholic and two non-alcoholic ingredients. Since order of arrangement does not matter, this is a combination problem.

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 ways these ingredients can be combined into a drink can be determined by the Multiplication Principle. The Multiplication Principle tells us that the number of ways independent events can occur together can be determined by multiplying together the number of possible outcomes for each event.

The number of possible drinks is

= 10 * 10
= 100
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by BTGmoderatorRO » Sun Sep 03, 2017 11:23 am
this is a combination problem,
and the formula is given as
C(n,r)=n!/ (r!(n-r!))

since we are having two ingredients which are alcoholic and non-alcoholic.
we will find their combination separately,

first let find that of non-alcoholic (na),
using the combination formula, C(n,r)=n!/ (r!(n-r!))
where n=5 and r=2
C(5,2)na= 5!/ (2! (5-2!))
5!=5*4*3*2*1= 120
2!=2*1=2
3!=3*2*1=6
C(5,2)= 120/ (2*6)
=120/12 =10

now, let us find the combination for the alcoholic(a)
C(n,r)a=n!/ (r!(n-r!))
n=5 and r=2
C(5,2)= 5!/ (2! (5-2!))
C(5,2)= 120/ (12)
=10

so we will apply the multiplication rule to determine the number of ways the ingredients can combine together.
C(5,2)na * C(5,2)a = 10*10= 100
so, for this question, the correct option is A.
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by [email protected] » Sun Sep 03, 2017 11:42 am
Hi All,

The Combination Formula is perfect for this question (as the various explanations have shown). When dealing with larger numbers, it often helps to 'simplify' these types of calculations (instead of multiplying out every individual calculation).

For example, with 5c2, you have:

(5)(4)(3)(2)(1)/(2)(1)(3)(2)(1)

You should notice that the (3)(2)(1) can be 'cancelled out' of both the numerator and denominator, leaving:

(5)(4)/(2)(1) = 20/2 = 10

Many of the 'shortcuts' that you'll be able to take advantage of in the Quant section are based on 'reorganizing' information that you've been given, so you should be on the lookout to save time (even if it's in small increments), by examining how you approach each prompt. It's possible that "your way" is not actually the fastest way.

GMAT assassins aren't born, they're made,
Rich
Contact Rich at [email protected]
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Re: Difficult Math Problem #61 - Combinations

by Scott@TargetTestPrep » Thu Feb 13, 2020 5:56 am
800guy wrote: ↑
Wed Nov 22, 2006 6:11 pm
 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

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