x2
Introduction
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Types of Arrangements
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Two More Methods
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Some Applications

# Some Examples of Counting Problems

These concepts are much easier to see when you're looking at specific examples. We'll look at some basic ones on this page and some more complex situations on the next one.

# Example 1

How many even four digit numbers are there where no digit appears twice?

First, let me clarify the question a little. We're looking for even numbers that have no repeating digits. So, 1234 would qualify but 1224 wouldn't because the number 2 is repeated twice.

This is a classic example of the multiplication principle in action where we can use the "drawing lines" method that I discussed in the last section. Since our number has four digits, there are four numbers that we have to chose so our diagram will have four lines:

To be even, our number has to end in 0, 2, 4, 6 or 8. (Remember that 0 is even.) That gives us five possibilities for the last position.

 4

At this point, we've used up one of the possible ten digits so there are nine possibilities left for the first spot.

 9 4

Because the digits can't repeat that leaves us with only eight possibilities for the second position and seven for the third spot.

 9 7 8 4

If we multiply those numbers together, we get 9 · 7 · 8 · 4 = 2,016 possible numbers.

# Example 3

How many numbers can be made from the numbers 1, 3, 5 and 8 assuming that the digits never repat within a number?

The best way to answer a question like this is to break it up into its componet parts: How many single digit numbers are there? How many two digit numbers are there? Etc. Once we've got all those numbers, we can apply the Addition Principle and get the final answer by adding them together.

 one digit numbers Clearly there are only 4 of these. 4 two digit numbers Any of the four available numbers can be used for the first digit which leaves three for the second digit. 4 · 3 = 12 three digit numbers In this case we have four options for the first spot, three for the second and two for the third. 4 · 3 · 2 = 24 four digit numbers In this case we have four options for the first spot, three for the second, two for the third and one for the last. 4 · 3 · 2 · 1 = 24

Adding those numbers together gives us 4 + 12 + 24 + 24 = 64 possible numbers.

# Example 2

How many ways can five men and three women be seated at a table if a woman has to be seated in the left hand chair and a man has to be seated in the right hand chair?

With this kind of question, you usually want to start by addressing the most restrictive rules first. In this case, a woman has to go in the left hand chair so there are only three options for that seat.

 3

Because the right hand chair has to have a man in it, there are only five possibilities for it.

 3 5

There are no restrictions on the remaining seats so any of the six remaining people can go in seat two.

 3 6 5

From here, it's a straightforward application of what we did in Example 1. There are five people remaining for the third seat, four for the fourth seat, etc.

 3 6 5 4 3 2 1 5

If we multiply those numbers together, we get 3 · 6 · 5 · 4 · 3 · 2 · 2 · 5 = 10,800 possible arrangements.

# Example 4

How many ways can five men and three women be seated in nine chairs?

If you look closely at the numbers you'll see that we've got a little challenge here: There are more seats than people. This may seem confusing at first but it's really just a straightforward application of what we've seen so far. Just let the empty seat count as a person. This gives us 9 possibilities for the first seat - it's either one of the five men, the three women or the empty seat.

 9

Now we just follow the pattern we've used in all the previous examples. There are eight possibilities for the second seat, seven for the third, etc.

 9 8 7 6 5 4 3 2 1

Multiplying those numbers together gives us 9 · 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 = 362,880 possible arrangements.

How many numbers can be made from the numbers 1, 3, 5 and 8?

This question is very similar to Example 3 except it doesn't have the restriction that digits can't be reused. That means that every digit will have four possibilities so our four situations become:

 one digit numbers 4 two digit numbers 4 · 4 = 16 three digit numbers 4 · 4 · 4 = 64 four digit numbers 4 · 4 · 4 · 4 = 256

Adding those numbers together gives us 4 + 16 + 64 + 256 = 340 possible numbers.

John has five different jackets, three different shirts and two pairs of shoes. How many different combinations can be made from his wardrobe?

This is a straightforward application of the Multiplication Principle.

total arrangements = (number of jackets) · (number of shirts) · (number of shoes) = 5 · 3 · 2 = 30

How many ways can three men and three women be seated in eight chairs if the two end chairs have to have a person in them?

In this case, the two end chairs need to be dealt with first. Any of the six people can go in the first chair which leaves five possibilities for the last chair.

 6 5

Now we've got four people left but six chairs. If we count each of the open chairs as a "person" then we have six possibilities for the second chair, five for the third, etc.

 6 6 5 4 3 2 1 5

That gives us a total of 6 · 6 · 5 · 4 · 3 · 2 · 1 · 5 = 21,600 arrangements.

Show that the number of ways that n objects can be arranged in r places, assuming that repetition is allowed, is nr

We can apply the Multiplication Principle here like we have been but it's going to be a little more "open ended" since we don't have an exact number of spaces. Our diagram will look this:

 ... 1 2 ... r-1 r

The numbers below the lines indicate which space the line represents. Since we have n items and repetition is allowed, each space has n options.

 n n ... n n 1 2 ... r-1 r

The Multiplication Principle tells us that the number of possibilities is the product of all the n's in the diagram. Since there are r n's that product would be "n multiplied by itself r times". Writing that with exponents, we would give us, nr.

How many different ways can someone be dealt five cards in a sequence, e.g. 8, 9, 10, Jack, Queen? (Hint: You can't have any card for the first position, e.g. a straight can't start with a queen.)

Let's think first about the first of our five cards. It has to be a ten or lower since, if it were higher than a ten, we would make it to the top of the suit (the king) and still have slots left to fill. There's a total of 40 cards that meet that description (the 10 cards times the 4 suits).

 40

Once we've picked the first card, the numbers on the remaining cards are fixed. E.g. if our first card was a seven then the second card has to be an eight, the third card has to be a nine, etc. The only options that we have at this point are the suit. The second card might have to be an eight but there are four eights so our second slot has four possibilities.

 40 4

The same argument applies to each of the three remaining slots so our final diagram will look like:

 40 4 4 4 4

Giving us 40 · 4 · 4 · 4 · 4 = 10,240 possible arrangements.