Combinational Analysis - GRE Subject Test: Math

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Question

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?

Answer

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:

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Question

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?

Answer

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.

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Question

There are possible flavor options at an ice cream shop.

$\textup{Quantity A: The number of possible combinations if three unique flavors are selected.}$

$\textup{Quantity B: The number of possible combinations if five unique flavors are selected.}$

Answer

When dealing with combinations, the number of possible combinations when selecting choices out of options is:

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

For Quantity A, the number of combinations is:

$C(10,3)=\frac{10!}{(10-3)!3!}$

For Quantity B, the number of combinations is:

$C(10,5)=\frac{10!}{(10-5)!5!}$

Quantity B is greater.

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Question

There are eight possible flavors of curry at a particular restaurant.

Quantity A: Number of possible combinations if four unique curries are selected.

Quantity B: Number of possible combinations if five unique curries are selected.

Answer

The number of potential combinations for selections made from possible options is

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

Quantity A:

$C(8,4)=\frac{8!}{(8-4)!4!}=70$

Quantity B:

$C(8,5)=\frac{8!}{(8-5)!5!}=56$

Quantity A is greater.

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Question

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.

Answer

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.

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Question

Quantity A: The number of potential combinations given two choices made from ten options.

Quantity B: The number of potential combinations given four choices made from twenty options.

Answer

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

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

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

Quantity A:

$C(10,2)=\frac{10!}{(10-2)!2!}=45$

Quantity B:

$C(20,4)=\frac{20!}{(20-4)!4!}=4845$

Quantity B is larger.

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Question

Quantity A: The number of combinations if five choices are made from ten options.

Quantity B: The number of combinations if two choices are made from twenty options.

Answer

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

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

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

Quantity A:

$C(10,5)=\frac{10!}{(10-5)!5!}=252$

Quantity B:

$C(20,2)=\frac{20!}{(20-2)!2!}=190$

Quantity A is greater.

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Question

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?

Answer

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.

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

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Question

A coach must choose starters from a team of players. How many ways can the coach choose the starters?

Answer

Step 1: We need to read the question carefully. Order does not matter here.

Step 2: Order does not matter, so we need to use Combination.

Step 3: The combination formula is $nCr=\frac{n!}{r!(n-r)!}$ .

Step 4: We need to find the value of and .

The value of is how many players the coach can choose from, so .
The value of is how many players that the coach can choose at one time, so .

Step 5: Plug in the values of n and r into the equation in step 2:

$nCr=\frac{11!}{5!(11-5)!}$

Step 6. Simplify the equation in step 5. The "!" means that I multiply that number by every other number below until 1.

$nCr=\frac{11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{5\cdot 4\cdot 3\cdot 2\cdot 1\cdot 6 \cdot5\cdot 4\cdot 3\cdot 2\cdot 1}$

Step 7: Cross out any terms that are on both the top and the bottom. We see is on top and bottom.

$nCr=\frac{11\cdot 10\cdot 9\cdot 8\cdot 7}{5\cdot4\cdot3\cdot2\cdot1}$

Step 8: Cross out $5\cdot2$ in the denominator with in the numerator. Rewrite.

$nCr=\frac{11\cdot 9\cdot 8\cdot 7}{4\cdot 3\cdot 1}$

Step 9: Divide in the numerator by in the denominator.

$nCr=\frac{11\cdot 9\cdot 2\cdot 7}{3\cdot 1}$

Step 10: Divide in the numerator by in the denominator.

$nCr=11\cdot 3\cdot2\cdot7$

Step 11: Multiply the right side

There are 462 ways that the coach can choose 5 players out of 11 players on the bench.

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Question

Find .

Answer

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,

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

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Question

Find .

Answer

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,

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

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

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

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Question

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

Answer

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.

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

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Question

Evaluate $_{5}C_{3}$ .

Answer

Evaluate $_{5}C_{3}$ is asking to calculate the combination of five objects when choosing three of them.

$_{5}C_{3} = \frac{5!}{(5-3)! 3!}$

$_{5}C_{3} = \frac{5!}{(5-3)! 3!} = \frac{5\times 4\times 3\times 2\times 1}{2\times 1\times3 \times 2\times 1}$

or $3\times2\times1$ cancels out.

$_{5}C_{3} = \frac{5\times 4 }{2\times 1}$

$\frac{20}{2} = 10$

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Question

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

Answer

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...

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Question

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

Answer

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: .

Step 4: Determine how many numbers I can put in the third and final digit...

I can put numbers in the third slot..

I had numbers at the start, and then I removed of them. .

Step 5: Multiply how many numbers can go in the first, second, and third spot..

$8\cdot8\cdot7=448$ .

There are a total of non-repetitive three-digit numbers that can be formed.

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Question

Six people run in a race, in how many different orders can they finish?

Answer

This problem is solved by knowing that we have six options for first place, five options for second place, and so on.

Which means

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Question

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?

Answer

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..

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

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Question

Simplify the following expression:

$\frac{12!}{10!-4!}$

Answer

Recall that ! means facotrial in math. This means we multiply the number by all positive integers less than itself. In other words, this...

$\frac{12!}{10!-4!}$

Becomes

$\frac{1\cdot 2\cdot3\cdot 4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10\cdot11\cdot12}{(1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10)-(1\cdot2\cdot3\cdot 4)}$

This is a great job for a calculator, which yields:

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Question

Evaluate:

Answer

Step 1: We need to define factorial (!). The operation (!) means that you multiply the number next to the ! by the next consecutive number down to 1.

Step 2: Expand the equation:

$5!=5\cdot4\cdot3\cdot2\cdot1=120$
$3(4\cdot 3\cdot 2\cdot 1)=3(24)=72$

Step 3: Evaluate the expression:

The answer to the expression is .

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Question

$\frac{5!}{3!-2!}$

Answer

is the product of consecutive numbers 1 through

$\frac{5\times 4\times 3\times 2\times 1}{(3\times 2\times 1)-(2\times 1)} =$

$\frac{120}{6-2} = \frac{120}{4} = 30$

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