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Statistics - II (Probability, Combinations and Permutations)

# GED Mathematical Reasoning: Combinations

• A combination is a way of selecting items from a collection, such that the order of selection does not matter.

Example 1

Five people are running for city council: Mary, Barb, Don, Harry, and Chris. Given that there are two seats available, how many combinations are possible if the order does not matter?

One strategy for doing so is to make a list. I’ll use the first letter of each person’s name to streamline the process, meaning I’ll use “M” for Mary, “B” for Barb, and so on.

M = Mary

B = Barb

D = Don

H = Harry

C = Chris

Mary can be selected along with Barb, as represented by “MB.” Or Mary can be selected along with Don, as represented by “MD.”  Or Mary can be selected along with Harry, as represented by “MH.”  Or Mary can be selected along with Chris, which we denote with “MC.”

Now let’s jump to Barb and list all the combinations involving Barb.

Barb can be selected along with Mary, but we already have that combination listed as “MB” so we won’t list it again. Remember – since the order doesn’t matter, “BM” is the same as “MB.” From there, Barb can be selected along with Don, Barb can be selected along with Harry, or Barb can be selected along with Chris. We’ll represent those combinations with “BD,”  “BH,”  and “BC” respectively.

Moving on to Don, Don can be selected along with Mary or Barb but we already have those combinations listed so we won’t list them again. Don can be selected along with Harry or Chris, though, so we’ll list those combinations as “DH” and “DC”.

Finally, Harry can be selected along with Mary, Barb, or Don – which we have already accounted for. The last remaining combination is that Harry is selected along with Chris.

So as you can see based on our list, there are TEN possible combinations.

Another strategy for solving this problem is to create a table. Title each column with the names and let each row represent a possible combination.

Note that each row accounts for two people being selected. We could not add any more rows to our table without duplicating a previous row.

And since there are ten rows (not counting the one containing column titles), that means there are ten possible combinations for selecting 2 people for city council out of 5 people.

Example 2

A local café offers a lunch special for \$4.99 where the diner selects one meat dish, one side dish and one dessert. The meat dish options are: beef or chicken. The side dish choices are: green beans, potato salad or carrots. And for dessert, the options are: key lime pie and chocolate cake. How many ways can the diner order lunch?

In example two, the situation calls for us to calculate the number of ways to choose one item from each of three different groups. Note, however, that the order still doesn’t matter – a diner’s order will be the same no matter if the diner says the meat dish before the side dish or the dessert before the meat dish.

A visual way to determine the number of possible combinations is to use a tree diagram. Consider the diagram on the slide and notice that the main “trunk” represents the meat dish options. From each meat dish option is an “offshoot” representing the side dish options. And from each side dish option, there’s an “offshoot” representing the dessert options.

The number of combinations is equal to the number of branches on the tree. Counting them up along the far right side, we see that there are twelve combinations, meaning there are 12 different ways to order lunch.

There is another way to find the number of possible combinations when one item is chosen from each of multiple groups. Since there are two ways to select the meat dish, three ways to select the side dish, and two ways to select dessert, the number of combinations can be found by multiplying two times three times two. Two times three times two equals twelve, meaning there are 12 combinations or ways to order lunch.

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