In how many ways can 6 differently colored rose bushes be planted in a row that only has room for 4 bushes?

Explanation:

There are 6 ways to choose the first rose bush, 5 ways to choose the second, 4 ways to choose the third, and 3 ways to choose the fourth. In total there are 6 * 5 * 4 * 3 = 360 ways to arrange the rose bushes.

You are watching: How many different arrangements of 5 letters can be made from the 26 letters of the alphabet?

Explanation:

We need to arrange the 5 balls in 5 positions: _ _ _ _ _. The first position can be taken by any of the 5 balls. Then there are 4 balls left to fill the second position, and so on. Therefore the number of arrangements is 5! = 5 * 4 * 3 * 2 * 1 = 120.

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?

Explanation:

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: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?

Explanation:

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.

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.

Explanation:

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.

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?

Explanation:

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

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?

Explanation:

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

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

**Possible Answers:**

40,320

36

362,880

3,040

15,700

**Correct answer:**40,320

Explanation:

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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### Example Question #2 : How To Find The Greatest Or Least Number Of Combinations

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

**Possible Answers:**

None of the other answers

89,700

45,000

300!

44,850

**Correct answer:**

44,850

Explanation:

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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### Example Question #3 : How To Find The Greatest Or Least Number Of Combinations

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:**

760,400

100,000

530,600

250,000

358,800

**Correct answer:**

358,800

Explanation:

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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