Permutations - GRE Quantitative Reasoning

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Question

Jill is picking out outfits for a three-day weekend, one for Friday, one for Saturday, and one for Sunday.

Fortunately all of her clothes match together really well, so she can be creative with her options, though she's decided each outfit is going to be a combination of blouse, skirt, and shoes.

She'll be picking from ten blouses, twelve skirts, and eight pairs of shoes. How many ways could her weekend ensemble be lined up?

Answer

For this problem, order matters! Wearing a particular blouse on Friday is not the same as wearing it on Sunday. So that means that this problem will be dealing with permutations.

With selections made from potential options, the total number of possible permutations(order matters) is:

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

What we'll do is calculate the number of permutations for her blouses, skirts, and shoes seperately (determining how the Friday/Saturday/Sunday blouses/skirts/shoes could be decided), and then multiply these values.

Blouses:

$P(10,3)=\frac{10!}{(10-3)!}=720$

Skirts:

$P(12,3)=\frac{12!}{(12-3)!}=1320$

Shoes:

$P(8,3)=\frac{8!}{(8-3)!}=336$

Thus the number of potential outfit assignments is

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Question

Lisa is dressing warm for the winter. She'll be layering three shirts over each other, and two pairs of socks. If she has fifteen shirts to choose from, along with ten different kinds of socks, how many ways can she layer up?

Answer

Since the order in which Lisa layers up matters, we're dealing with permutations.

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

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

For her shirts:

$P(15,3)=\frac{15!}{(15-3)!}=2730$

For her socks:

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

Her total ensemble options is the product of these two results

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Question

Quantity A: The number of possible permutations when seven choices are made from ten options.

Quantity B: The number of possible permutations when five choices are made from eleven options.

Answer

With selections made from potential options, the total number of possible permutations(order matters) is:

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

Quantity A:

$P(10,7)=\frac{10!}{(10-7)!}=604800$

Quantity B:

$P(11,5)=\frac{11!}{(11-5)!}=55440$

Quantity A is greater.

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Question

Daisy wants to arrange four vases in a row outside of her garden. She has eight vases to choose from. How many vase arrangements can she make?

Answer

For this problem, since the order of the vases matters (red blue yellow is different than blue red yellow), we're dealing with permutations.

With selections made from potential options, the total number of possible permutations(order matters) is:

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

$P(8,4)=\frac{8!}{(8-4)!}=1680$

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Question

Abby, Bryan, Cindy, Doug, and Ernie are sitting on a bench. How many ways can I arrange their seating order?

Answer

Step 1: We need to identify how many seats there are on the bench. We have 5 names, so 5 seats.

Step 2: When 1 person sits in seat 1, he/she cannot sit in the next set, and so on.

Step 3: Let's work out the math...

Seat 1- 5 people can sit
Seat 2- 4 people can sit
Seat 3- 3 people can sit
Seat 4/5-2/1 people/person can sit

Total possibilities= $5\cdot4\cdot3\cdot2\cdot1=120$ . We can also say that

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Question

There are 12 boys in a football competition, the top 3 competitors are awarded with an trophy. How may possible groups of 3 are there for this competition?

Answer

This is a permutation. A permutation is an arrangement of objects in a specific order.

The formula for permutations is:

$_{n}P_{k} = \frac{n!}{(n-k)!} - n(n-1) (n-2)...(n-k+1)$

This is written as $_{n}P_{k}$

$_{12}P_{3} = 12\times11\times10 = 1,320$

There are possible groups of 3.

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Question

An ice cream shop has 23 flavors. Melissa wants to buy a 3-scoop cone with 3 different flavors, How many cones can she buy if order is important?

Answer

This is a permutation. A permutation is an arrangement of objects in a specific order.

The formula for permutations is:

$_{n}P_{k} = \frac{n!}{(n-k)!} - n(n-1) (n-2)...(n-k+1)$

This is written as $_{n}P_{k}$

$_{23}P_3=P(23,3)$ represents the number of permutations of 23 things taken 3 at a time.

$P(23,3) = 23\times22\times21 = 10,626$

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Question

Find the value of .

Answer

is asking to find the permutation of seven items when you want to choose five. When dealing with permutations, order matters.

A permutation is an arrangement of objects in a specific order.

The formula for permutations is:

$_{n}P_{k} = \frac{n!}{(n-k)!} - n(n-1) (n-2)...(n-k+1)$

This is written as

$\_nP_k\_7P_5=P(7,5) = 7\times6\times5\times4\times3 = 2,520$

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Question

Evaluate .

Answer

is asking to find the permutation of four items when you want to choose all four. When dealing with permutations, order matters.

A permutation is an arrangement of objects in a specific order.

The formula for permutations in this case will be,

or factorial.

$P(4,4) = 4\times3\times2\times1 = 24$

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Question

There are people at a family dinner. After the dinner is over, people shake hands with each other. How many handshakes were there between these people. Note: Once two people shake hands, they cannot shake hands again..

Answer

Step 1: A handshake MUST ALWAYS be between TWO people.

Step 2: Break down each person and who they can shake hands with:

Person can shake hands with: .
Person can shake hands with:
Person can shake hands with:
Person can shake hands with:
Person can shake hands with:
Person can shake hands with:
Person can shake hands with:
Person can shake hands with:
Person can shake hands with:
Person can shake hands with:
Person already shook everybody's hand..

Step 3: Count how many handshakes each person can make:

Person shakes hands times.
Person shakes hands times.
Person shakes hands times.
Person shakes hands times.
Person shakes hands times.
Person shakes hands times.
Person shakes hands times.
Person shakes hands times.
Person shakes hands times.
Person shakes hands time.
Person already shook everybody's hand.

Step 4: Add up the number of times each person shook hands:

There were handshakes made between these people.

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Question

How many three-digit numbers can I create from the set of numbers ?

Answer

Step 1: Identify if there are any restrictions to how the numbers can be made...

There are no restrictions, so we can have repeating numbers.

Step 2: Determine how many numbers can go in each slot..

First Slot: 7 choices
Second Slot: 7 Choices
Third Slot: 7 choices

Step 3: Multiply the choices for all three sets together:

$7\cdot7\cdot7\equiv7^3\equiv 343$

We can create different three-digit numbers...

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Question

How many non-repetitive three-digit numbers can I create from the set of numbers ?

Answer

Step 1: See if there are any restrictions..

We see we want only non-repetitive numbers...

Step 2: Find how many numbers can be put in each spot:

First Spot:
Second Spot:
Third Spot:

Step 3: Multiply the number of choices of each spot

$8\cdot6\cdot7=336$

I can create non-repetitive three-digit numbers...

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Question

How many ways can I arrange the letters in the word ?

Answer

Step 1: Count how many letters are in the word MISSISSIPPI...

There are 11 numbers.

Step 2: Find which letters repeat, and how many times it repeats:

times
times
times

Step 3: Use formula for arranging letters:

$\frac {(Total\Letters)!}{(Repeats)!}=\frac {11!}{2!4!4!}$

Step 4: Expand:

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

Step 5: Simplify Step 4:

$\frac {11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5}{4}=11\cdot10\cdot9\cdot2\cdot7\cdot6\cdot5=34650$

I can rearrange the letters in MISSISSIPPI times...

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Question

How many ways can I arrange the letters in the word CORRECT?

Answer

Step 1: Count how many letters are in the word CORRECT...

There are letters.

Step 2: Find any letters that repeat and how many times they repeat:

C (two times), R (two times)

Step 3: Find how many ways can I arrange the letters:

$\frac{7!}{2!2!}=\frac {7654321}{2121}=7653*2=1260$

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Question

people are at a farewell party. At the end of the night, each person shakes hands. How many handshakes are made?

NOTE: No two people can shake hands more than once.

Answer

Step 1: Determine how many people are there..
There are people.
Step 2: Determine how many people shake hands in a handshake...
people make one handshake.
Step 3: Determine how many handshakes can be made...
We have a restriction here, so we need to use permutation..

So, there will be handshakes.

$11P2=\frac {1110}{21}=11*5=55$ handshakes

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Question

In how many ways can I rearrange the letters in the word "ANACONDA"?

Answer

Step 1: Count how many letters are in the word ANACONDA...

There are 8 letters.

Step 2: Count how many repeats of any letters (if any)..

There are A's and N's.

Step 3: Find how many ways I can rearrange...

$\frac {8!}{3!2!}=\frac {87654321}{32121}...$

$...=\frac {87654}{2}=87652=3360$

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Question

How many ways can I arrange the letters in the word ?

Answer

Step 1: Count how many numbers are in the word...

There are letters.

Step 2: Count the number of repeated letters...

There are T's

There are O's

There are R's

Step 3: To find how many ways I can arrange the letters, take the factorial of the total number of letters and divide it by the factorial of how many times a certain letter repeats...

So, $\frac {12!}{2!\times2!\times2!}$

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Question

Lisa is dressing warm for the winter. She'll be layering three shirts over each other, and two pairs of socks. If she has fifteen shirts to choose from, along with ten different kinds of socks, how many ways can she layer up?

Answer

Since the order in which Lisa layers up matters, we're dealing with permutations.

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

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

For her shirts:

$P(15,3)=\frac{15!}{(15-3)!}=2730$

For her socks:

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

Her total ensemble options is the product of these two results

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Question

Quantity A: The number of possible permutations when seven choices are made from ten options.

Quantity B: The number of possible permutations when five choices are made from eleven options.

Answer

With selections made from potential options, the total number of possible permutations(order matters) is:

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

Quantity A:

$P(10,7)=\frac{10!}{(10-7)!}=604800$

Quantity B:

$P(11,5)=\frac{11!}{(11-5)!}=55440$

Quantity A is greater.

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Question

Jill is picking out outfits for a three-day weekend, one for Friday, one for Saturday, and one for Sunday.

Fortunately all of her clothes match together really well, so she can be creative with her options, though she's decided each outfit is going to be a combination of blouse, skirt, and shoes.

She'll be picking from ten blouses, twelve skirts, and eight pairs of shoes. How many ways could her weekend ensemble be lined up?

Answer

For this problem, order matters! Wearing a particular blouse on Friday is not the same as wearing it on Sunday. So that means that this problem will be dealing with permutations.

With selections made from potential options, the total number of possible permutations(order matters) is:

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

What we'll do is calculate the number of permutations for her blouses, skirts, and shoes seperately (determining how the Friday/Saturday/Sunday blouses/skirts/shoes could be decided), and then multiply these values.

Blouses:

$P(10,3)=\frac{10!}{(10-3)!}=720$

Skirts:

$P(12,3)=\frac{12!}{(12-3)!}=1320$

Shoes:

$P(8,3)=\frac{8!}{(8-3)!}=336$

Thus the number of potential outfit assignments is

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