Combinations

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GRE Quantitative Reasoning › Combinations

Questions 1 - 10

Mohammed is being treated to ice cream for his birthday, and he's allowed to build a three-scoop sundae from any of the thirty-one available flavors, with the only condition being that each of these flavors be unique. He's also allowed to pick different toppings of the available , although he's already decided well in advance that one of them is going to be peanut butter cup pieces.

Knowing these details, how many sundae combinations are available?

Explanation

Because order is not important in this problem (i.e. chocolate chip, pecan, butterscotch is no different than pecan, butterscotch, chocolate chip), it is dealing with combinations rather than permutations.

The formula for a combination is given as:

$C(n,k)=\frac{n!}{(n-k)!k!}$

where is the number of options and is the size of the combination.

For the ice cream choices, there are thirty-one options to build a three-scoop sundae. So, the number of ice cream combinations is given as:

$C(31,3)=\frac{31!}{(31-3)!3!}=4495$

Now, for the topping combinations, we are told there are ten options and that Mohammed is allowed to pick two items; however, we are also told that Mohammed has already chosen one, so this leaves nine options with one item being selected:

$C(9,1)=\frac{9!}{(9-1)!1!}=9$

So there are 9 "combinations" (using the word a bit loosely) available for the toppings. This is perhaps intuitive, but it's worth doing the math.

Now, to find the total sundae combinations—ice cream and toppings both—we multiply these two totals:

How many ways can a coach choose players to play on the field out of a bench of players?

$\frac {13!}{7!-6!}$

Explanation

Step 1: Read the question carefully. Look for hints of restrictions..

There are no order in which players can be chosen, which goes against the definition of Permutation. Permutation is the arrangement of objects by way of order.. If it's not permutation, it's Combination.

Step 2: Write what we know down..

Total Players =
Choosing # of players =..

Step 3: Plug in the numbers to the formula: ..

We ger 13C6.

There is no need to evaluate this expression...

Find .

Explanation

There are two types of statistical calculations that are used when dealing with ordering a number of objects. When the order does not matter it is known as a combination and denoted by a C.

$C(n,k)=\frac{n!}{k!(n-k)!}$

Thus the formula for this particular combination is,

$C (8,5) = P \frac{(8,5)}{5!}$

$\frac{8\times 7\times 6\times 5\times 4}{5\times 4\times 3\times 2\times 1} =$

The $5 \times 4$ will cancel out because it is in the numerator and denominator,

$\frac{8\times 7\times 6}{3\times 2\times 1} = \frac{336}{6} = 56$ .

A coach of a baseball team needs to choose players out of a total players in the team. How many ways can the coach choose 9 players?

Explanation

Step 1: Recall the combination formula...

$_nC_r=\frac {n!}{r!(n-r)!}$

Step 2: Find and from the question..

Step 3: Plug in the values into the formula above..

$\frac {17!}{9!(17-9)!}=\frac {17!}{9!\times8!}=24,310$

Find .

Explanation

There are two types of statistical calculations that are used when dealing with ordering a number of objects. When the order does not matter it is known as a combination and denoted by a C.

$C(n,k)=\frac{n!}{k!(n-k)!}$

Thus the formula for this particular combination is,

$C (10,4) = P \frac{(10,4)}{4!}$

$\frac{10\times 9\times 8\times 7}{ 4\times 3\times 2\times 1} =$

$\frac{90\times 56}{12\times 2} = \frac{5040}{24} = 210$

Quantity A: The number of possible combinations if four unique choices are made from ten possible options.

Quantity B: The number of possible permutations if two unique choices are made from ten possible options.

Quantity A is greater.

Quantity B is greater.

The two quantities are equal.

The relationship cannot be established.

Explanation

For choices made from possible options, the number of potential combinations (order does not matter) is

$C(n,k)=\frac{n!}{(n-k)!k!}$

And the number of potential permutations (order matters) is

$P(n,k)=\frac{n!}{(n-k)!}$

Quantity A:

$C(10,4)=\frac{10!}{(10-4)!4!}=210$

Quantity B:

$P(10,2)=\frac{10!}{(10-2)!}=90$

Quantity A is greater.

Rachel is buying ice cream for a sundae. If there are twelve ice cream choices, how many scoops will give the maximum possible number of unique sundaes?

Explanation

Since in this problem the order of selection does not matter, we're dealing with combinations.

With selections made from potential options, the total number of possible combinations is

$C(n,k)=\frac{n!}{(n-k)!k!}$

In terms of finding the maximum number of combinations, the value of should be

$k_{max}=\frac{n_{even}}{2};k_{max}=\frac{n_{odd}\pm 1}{2}$

Since there are twelve options, a selection of six scoops will give the maximum number of combinations.

$C(12,6)=\frac{12!}{(12-6)!6!}=924$

If there are students in a class and people are randomly choosen to become class representatives, how many different ways can the representatives be chosen?

Explanation

To solve this problem, we must understand the concept of combination/permutations. A combination is used when the order doesn't matter while a permutation is used when order matters. In this problem, the two class representatives are randomly chosen, therefore it doesn't matter what order the representative is chosen in, the end result is the same. The general formula for combinations is $C(n,k)=\frac{n!}{(n-k)!k!}$ , where is the number of things you have and is the things you want to combine.

Plugging in choosing 2 people from a group of 20, we find

$C(20,2)=\frac{20!}{(20-2)!2!}=190$ . Therefore there are a different ways to choose the class representatives.

Six points are located on a circle. How many line segments can be drawn?

Explanation

There are two types of statistical calculations that are used when dealing with ordering a number of objects. When the order does not matter it is known as a combination and denoted by a C.

$C(n,k)=\frac{n!}{k!(n-k)!}$

Thus the formula for this particular combination is,

$C=\frac{P(6,2)}{2!}$

There are 2 points on each line segment.

$\frac{6\times 5}{ 2\times 1} = \frac{30}{2} = 15$

How many ways can I get non-repetitive three-digit numbers from the numbers: ?

Explanation

Step 1: Count how many numbers I can use..

I can use 9 numbers.

Step 2: Determine how many numbers I can put in the first digit of the three-digit number..

I can put numbers in the first spot. I cannot put in the first slot because the number will not be a three-digit number.

Step 3: Determine how many numbers I can put in the second digit..

I can also put numbers in the second spot. Here's the reason why it's still :

Let's say I choose 2 for the first number. I will take out of my set. I had numbers in my set..If i take a number out, I still have numbers left. These numbers are: .