### All GRE Math Resources

## Example Questions

### Example Question #1 : How To Find The Greatest Or Least Number Of Combinations

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?

**Possible Answers:**

32,760

105

250

210

**Correct answer:**

105

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.

### Example Question #5 : Permutation / Combination

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.

**Possible Answers:**

Quantity A is greater.

Quantity B is greater.

The relationship cannot be determined from the information given.

The quantities are equal.

**Correct answer:**

Quantity B is greater.

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.

### Example Question #4 : Permutation / Combination

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?

**Possible Answers:**

None of these

240

1145

720

120

**Correct answer:**

120

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

### Example Question #7 : Permutation / Combination

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?

**Possible Answers:**

231

365580800

456426360

219

619652800

**Correct answer:**

456426360

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.

### Example Question #11 : Permutation / Combination

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

**Possible Answers:**

**Correct answer:**40,320

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.

### Example Question #12 : Permutation / Combination

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

**Possible Answers:**

300!

45,000

89,700

44,850

None of the other answers

**Correct answer:**

44,850

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

### Example Question #12 : Permutation / Combination

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.

**Possible Answers:**

530,600

358,800

100,000

250,000

760,400

**Correct answer:**

358,800

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

### Example Question #13 : Permutation / Combination

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

**Possible Answers:**

1020

1400

6500

5040

4230

**Correct answer:**

5040

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.

### Example Question #14 : Permutation / Combination

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

**Possible Answers:**

1820

2184

43,680

16

4

**Correct answer:**

1820

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: and divide by number of spots open

### Example Question #15 : Permutation / Combination

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?

**Possible Answers:**

**Correct answer:**

The customer has 3 choices on meat, 6 choices on side, and 5 choices on drink. This gives a total of choices for the meal.

### All GRE Math Resources

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