All GRE Subject Test: Math Resources
Example Questions
Example Question #1 : Combinations
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?
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 , 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
. Therefore there are a different ways to choose the class representatives.
Example Question #1 : Combinations
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.
Quantity A is greater.
The relationship cannot be determined.
Quantity B is greater.
The two quantities are equal
Quantity A is greater.
The number of potential combinations for selections made from possible options is
Quantity A:
Quantity B:
Quantity A is greater.
Example Question #23 : Permutation / Combination
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 B is greater.
Quantity A is greater.
The relationship cannot be established.
The two quantities are equal.
Quantity A is greater.
For choices made from possible options, the number of potential combinations (order does not matter) is
And the number of potential permutations (order matters) is
Quantity A:
Quantity B:
Quantity A is greater.
Example Question #2 : Combinations
There are possible flavor options at an ice cream shop.
When dealing with combinations, the number of possible combinations when selecting choices out of options is:
For Quantity A, the number of combinations is:
For Quantity B, the number of combinations is:
Quantity B is greater.
Example Question #21 : Permutation / Combination
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.
Quantity A is larger.
The two quantities are equal.
Quantity B is larger.
The relationship cannot be determined.
Quantity B is larger.
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
Quantity A:
Quantity B:
Quantity B is larger.
Example Question #22 : Permutation / Combination
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.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined.
Quantity A is greater.
Quantity A is greater.
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
Quantity A:
Quantity B:
Quantity A is greater.
Example Question #72 : Other Topics
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?
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
In terms of finding the maximum number of combinations, the value of should be
Since there are twelve options, a selection of six scoops will give the maximum number of combinations.
Example Question #1 : Combinations
A coach must choose starters from a team of players. How many ways can the coach choose the starters?
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 .
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:
Step 6. Simplify the equation in step 5. The "!" means that I multiply that number by every other number below until 1.
Step 7: Cross out any terms that are on both the top and the bottom. We see is on top and bottom.
Step 8: Cross out in the denominator with in the numerator. Rewrite.
Step 9: Divide in the numerator by in the denominator.
Step 10: Divide in the numerator by in the denominator.
Step 11: Multiply the right side
There are 462 ways that the coach can choose 5 players out of 11 players on the bench.
Example Question #2 : Combinations
How many ways can a coach choose players to play on the field out of a bench of players?
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...
Example Question #571 : Gre Subject Test: Math
Find .
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.
Thus the formula for this particular combination is,
The will cancel out because it is in the numerator and denominator,
.
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