All GRE Subject Test: Math Resources
Example Questions
Example Question #21 :Permutation / Combination
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 pickdifferent 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?
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:
whereis the number of options andis 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:
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:
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.
Example Question #330 :Arithmetic
If there arestudents in a class andpeople are randomly choosen to become class representatives, how many different ways can the representatives be chosen?
所以lve 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, whereis the number of things you have andis the things you want to combine.
Plugging in choosing 2 people from a group of 20, we find
. Therefore there are adifferent ways to choose theclass representatives.
Example Question #21 :Permutation / Combination
There are eight possible flavors of curry at a particular restaurant.
Quantity A: Number of possible combinations if four unique curries are selected.
量B:如果fiv的可能的组合e unique curries are selected.
The two quantities are equal
The relationship cannot be determined.
Quantity B is greater.
Quantity A is greater.
Quantity A is greater.
The number of potential combinations forselections made frompossible options is
Quantity A:
Quantity B:
Quantity A is greater.
Example Question #22 :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.
The relationship cannot be established.
Quantity A is greater.
The two quantities are equal.
Quantity B is greater.
Quantity A is greater.
Forchoices made frompossible 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 #1 :Combinations
There arepossible flavor options at an ice cream shop.
When dealing with combinations, the number of possible combinations when selectingchoices out ofoptions is:
For Quantity A, the number of combinations is:
For Quantity B, the number of combinations is:
Quantity B is greater.
Example Question #24 :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.
The relationship cannot be determined.
The two quantities are equal.
Quantity B is larger.
Quantity A is larger.
Quantity B is larger.
Since in this problem we're dealing with combinations, the order of selection does not matter.
Withselections made frompotential options, the total number of possible combinations is
Quantity A:
Quantity B:
Quantity B is larger.
Example Question #26 :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.
The relationship cannot be determined.
Quantity A is greater.
The two quantities are equal.
Quantity B is greater.
Quantity A is greater.
Since we're dealing with combinations in this problem, the order of selection does not matter.
Withselections made frompotential options, the total number of possible combinations is
Quantity A:
Quantity B:
Quantity A is greater.
Example Question #34 :Permutation / Combination
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.
Withselections made frompotential options, the total number of possible combinations is
In terms of finding the maximum number of combinations, the value of应该是
Since there are twelve options, a selection of six scoops will give the maximum number of combinations.
Example Question #71 :Probability & Statistics
A coach must choosestarters from a team ofplayers. 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 ofand.
The value ofis how many players the coach can choose from, so.
The value ofis 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 seeis on top and bottom.
Step 8: Cross outin the denominator within the numerator. Rewrite.
Step 9: Dividein the numerator byin the denominator.
Step 10: Dividein the numerator byin 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 #72 :Probability & Statistics
How many ways can a coach chooseplayers to play on the field out of a bench ofplayers?
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...
Certified Tutor
Certified Tutor