In this lesson, we will practice solving various permutation and combination problems using permutation and combination formulas. We can continue our practice when we take a quiz at the end of the lesson.
Permutations and Combinations
Both permutations and combinations are groups or arrangements of objects. However, there is one big difference between them. When dealing with combinations, the order of the objects is insignificant, whereas in permutations the order of the objects makes a difference.
For example, assume you have 10 coins in your pocket and you take 5 out, a dime, 2 quarters, a nickel and a penny. If I said you grabbed those same 5 coins, but I said you grabbed 2 quarters, a nickel, a penny, and a dime, it is still the same group of coins. That is, the order I name them in is insignificant. Therefore, the coins are a combination of 5 of 10 coins.
Now consider the scenario where we are talking about 5 finishers in a race, runners A, B, C, D, and E. If I tell you they crossed the line in the order A, B, C, D, E, this would be different than if I told you they crossed the line in the order C, B, A, E, D. Thus, the order makes a difference, so the order in which the 5 runners finish is a permutation of the 5 runners.
Solving Permutation and Combination Problems
There are two questions you have to answer before solving a permutation/combination problem.
1.) Are we dealing with permutations or combinations? In other words, does order matter?
2.) Is repetition allowed? In other words, can we name an object more than once in our permutation or combination?
Once we have answered these questions, we use the appropriate formula to solve the problem. Those formulas are shown below.
Let's look at some examples to get comfortable solving these types of problems.
1.) How many distinct ways can the letters of the word FRIEND be arranged?
Solution: We will start off by determining if we're dealing with permutations or combinations. To do this, consider the arrangement FRIEND and the arrangement DIREFN. We see that these represent two different arrangements, so the order of the letters make a difference. Thus, we're dealing with permutations. Now we must consider if repetition is allowed. When we create an arrangement of the letters, we put one letter in the first spot, then the second spot, and so on until all 6 spots are filled. Once we put a letter in a spot, we can't put it in another spot, because it has already been used. Therefore, repetition is not allowed. We know that we are dealing with a permutation of 6 objects (letters) where repetition is not allowed. Therefore, we use formula n! We plug 6 in for n in the formula to get the following.
6! = 6 * 5 * 4 * 3 * 2 * 1 = 720
We see that there are 720 ways to arrange the letters in the word FRIEND.
2.) Assume you must select 3 people from a group of 20 people. How many ways are there to do this?
Solution: If we select 3 people, say George, Frank, and Jessica, it is still the same group if we said we selected Frank, Jessica, and George. Therefore, order is insignificant, so we are dealing with combinations. Once we have selected a person, we can't select them again since they are already in the group. Therefore, repetition is not allowed. Based on this information, we use the formula nCr = n! / r!(n - r)!, where we plug in 3 for r and 20 for n. Doing so gives the following.
Solution: Let's consider the 3-digit number 702 formed using 3 of the 5 digits. If we change the order to 207, this represents a different number. Therefore, order makes a difference, so we are dealing with permutations. Now, we can make the number 777 using the 5 given digits, because 7 is one of the digits. Therefore, repetition is allowed. We see we have a permutation of 3 objects from 5 objects, where repetition is allowed. Therefore, we use the formula n ^ r, where n = 5 and r = 3. Plugging these values in gives:
5 ^ 3 = 5*5*5 = 125
Therefore, we can make 125 3-digit numbers from the numbers 3, 7, 0, 2, and 9.
4.) There are 9 bikers in a bike race. In how many ways can the first three finishers come in?
Solution: Consider bikers A, B, and C finishing in that order. Now consider those same bikers finishing in the order C, A, B. This is a different order of finishers, so we see that order makes a difference, so we're dealing with permutations. Now, consider biker A coming in first. That biker has now finished, so the biker can't come in second, third, etc. Thus, repetition is not allowed. We are dealing with permuting 3 of 9 objects, where repetition is not allowed. Therefore, we use the formula nPr = n! / (n - r)!, and we plug 3 in for r and 9 in for n.
9! / (9 - 3)! = 9! / 6! = 9*8*7 = 504
There are 504 ways for the top three finishers to come in during a race with 9 participants.
When solving permutation and combination problems, we first ask ourselves two questions. First, are we dealing with permutations, where order matters or combinations, where order doesn't matter? Second, is repetition is allowed? Based on the answers to these questions, we use the appropriate formula to solve the problem posed.
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