How to find the greatest or least number of combinations - GRE Quantitative Reasoning

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Question

What is the minimum amount of handshakes that can occur among fifteen people in a meeting, if each person only shakes each other person's hand once?

Answer

This is a combination problem of the form “15 choose 2” because the sets of handshakes do not matter in order. (That is, “A shakes B’s hand” is the same as “B shakes A’s hand.”) Using the standard formula we get: 15!/((15 – 2)! * 2!) = 15!/(13! * 2!) = (15 * 14)/2 = 15 * 7 = 105.

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Question

There are 20 people eligible for town council, which has three elected members.

Quantity A

The number of possible combinations of council members, presuming no differentiation among office-holders.

Quantity B

The number of possible combinations of council members, given that the council has a president, vice president, and treasurer.

Answer

This is a matter of permutations and combinations. You could solve this using the appropriate formulas, but it is always the case that you can make more permutations than combinations for all groups of size greater than one because the order of selection matters; therefore, without doing the math, you know that B must be the answer.

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Question

The President has to choose from six members of congress to serve on a committee of three possible members. How many different groups of three could he choose?

Answer

This is a combinations problem which means order does not matter. For his first choice the president can choose from 6, the second 5, and the third 4 so you may think the answer is 6 * 5 * 4, or 120; however this would be the answer if he were choosing an ordered set like vice president, secretary of state, and chief of staff. In this case order does not matter, so you must divide the 6 * 5 * 4 by 3 * 2 * 1, for the three seats he’s choosing. The answer is 120/6, or 20.

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Question

Joe has a set of 10 books that he hasn't yet read. If he takes 3 of them on vacation, how many possible sets of books can he take?

Answer

He can choose from 10, then 9, then 8 books, but because order does not matter we need to divide by 3 factorial

(10 * 9 * 8) ÷ (3 * 2 * 1) = 720/6 = 120

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Question

How many different license passwords can one make if said password must contain exactly 6 characters, two of which are distinct numbers, another of which must be an upper-case letter, and the remaining 3 can be any digit or letter (upper- or lower-case) such that there are no repetitions of any characters in the password?

Answer

Begin by considering the three "hard and fast conditions" - the digits and the one upper-case letter. For the first number, you will have 10 choices and for the second 9 (since you cannot repeat). For the captial letter, you have 26 choices. Thus far, your password has 10 * 9 * 26 possible combinations.

Now, given your remaining options, you have 8 digits, 25 upper-case letters, and 26 lower-case letters (i.e. 59 possible choices). Since you cannot repeat, you will thus have for your remaining choices 59, 58, and 57 possibilities.

Putting all of this together, you have: 10 * 9 * 26 59 58 * 57 or 456426360 choices.

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Question

A restaurant has a meal special that allows you to choose one of three salads, one of five sandwiches, and two of fifteen side dishes. How many possible combinations are there for the meal?

Answer

Although this is a permutation style problem, we have to be careful regarding the last portion (i.e. the side dishes). We know that our meal will have:

(3 possible salads) * (5 possible sandwiches) * (x possible combinations of side dishes).

We must ascertain how many side dishes can be selected. Now, it does not matter what order we put together the side dishes, so we have to use the combinations formula:

c(n,r) = (n!) / ((n-r)! * (r!))

Plugging in, we get: c(15,2) = 15! / (13! * 2!) = 15 * 14 / 2 = 15 * 7 = 105

Using this in the equation above, we get: 3 * 5 * 105 = 1575

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Question

In how many different orders can 8 players sit on the basketball bench?

Answer

Using the Fundamental Counting Principle, there would be 8 choices for the first player, 7 choices for the second player, 6 for the third, 5 for the fourth, and so on. Thus, 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 or 8! = 40, 320.

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Question

There are 300 people at a networking meeting. How many different handshakes are possible among this group?

Answer

Since the order of persons shaking hands does not matter, this is a case of computing combinations. (i.e. It is the same thing for person 1 to shake hands with person 2 as it is for person 2 to shake hands with person 1.)

According to our combinations formula, we have:

300! / ((300-2)! * 2!) = 300! / (298! * 2) = 300 * 299 / 2 = 150 * 299 = 44,850 different handshakes

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Question

What is the number of possible 4-letter words that can be made from the 26 letters in the alphabet, where all 4 letters must be different?

Assume non-sensical words count, i.e. "dnts" would count as a 4-letter word for our purposes.

Answer

This is a permutation of 26 letters taken 4 at a time. To compute this we multiply 26 * 25 * 24 * 23 = 358,800.

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Question

10 people want to sit on a bench, but the bench only has 4 seats. How many arrangements are possible?

Answer

The first seat can be filled in 10 ways, the second in 9 ways, the third in 8 ways, and the fourth in 7 ways. So the number of arrangements = 10 * 9 * 8 * 7 = 5040.

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Question

There are 16 members of a club. 4 will be selected to leadership positions. How many combinations of leaders are possible?

Answer

With permutations and combinations, you have to know if the order people are selected matters or not. If not, like in this case, you must take the number of people and positions available: $\dpi{100} \small 16 \times 15 \times \times 14 \times 13$ and divide by number of spots open $\dpi{100} \small 4 \times 3 \times 2 \times 1 = 1820$

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Question

How many different committees of 3 people can be formed from a group of 7 people?

Answer

There are $\dpi{100} \small 7\times 6\times 5=210$ different permutations of 3 people from a group of seven (when order matters). There are $\dpi{100} \small 3\times 2=6$ possible ways to arrange 3 people. Thus when order doesn't matter, there can be $\dpi{100} \small \frac{210}{6}=35$ different committees formed.

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Question

A restaurant serves its steak entree cooked rare, medium, or well done. The customer has the choice of salad or soup, with one of two salads or one of 4 soups. The customer also chooses between one of three soft drinks as well as water or milk. How many unique variations are there to the entire steak dinner of

steak + soup/salad + drink?

Answer

The customer has 3 choices on meat, 6 choices on side, and 5 choices on drink. This gives a total of $\dpi{100} \small 3 \times 5\times 6=90$ choices for the meal.

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Question

From a group of 8 students, 3 are attending a meeting.

Quantity A: The number of different groups that could attend among the 8 students

Quantity B: 336

Answer

To solve this problem, you would need to utilize the combination formula, which is $\dpi{100} \small C=\frac{n!}{r!(n-r!)}$ .

$\dpi{100} \small C$ is the number of possibilities, $\dpi{100} \small n$ is the number of students, and $\dpi{100} \small r$ is the students attending the meeting. Thus, $\dpi{100} \small \frac{8!}{3!5!}=56$ .

336 would be the result of computing the permutation, which would be incorrect in this case.

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Question

Michael owns 10 paintings. Michael would like to hang a single painting in each of five different rooms. How many different ways are there for Michael to hang 5 of his 10 paintings?

Answer

This problem involves the permutation of 10 items across 5 slots. The first slot (room) can have 10 different paintings, the second slot can have 9 (one is already in the first room), the third slot can have 8 and so on. The number of possibilities is obtained by multiplying the number of possible options in each of the 5 slots together, which here is $\dpi{100} \small 10\times 9\times 8\times 7\times 6=30,240$ .

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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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