Mathematics
OpenStudy (anonymous):

A committee of four congressmen will be selected from a group 7 democrats and 3 republicans. find the number of different committees containing A).exactly one democrat B).exactly three democrats

OpenStudy (anonymous):

When answering "how many combinations" problems, we imagine we are placing the items (people in this case) into positions. To get the answer, we then multiply the possible ways to place weach position. So... A) Exactly one democrat: Since we need exactly 1 democrat, there are 7 possible people to choose from. Now because the committee is to have 4 people, and we have 1 democrat, the remaining 3 people would be republicans. There is only one way to pick those 3 republicans since that's all there is. So, we have 7 ways to pick the democrat and 1 way to pick the republicans, giving us $7\times1=7$ possible combinations. B) Exactly three democrats: This is a little harder because we run the risk of "double counting" committee members. Since we need 4 people total, and 3 are democrats, that leaves the last member as a republican. We'll deal with them last. To select the three democrats, we pick them one at a time. The first one, we have 7 possible choices, then 6 for the second, and 5 for the third (since we can't pick the same person twice). You would think that that would mean we have $7\times6\times5=210$choices. However, selecting person A, B, then C gives the same group as selecting perscn C, B, then A. So we have to reduce this. Once we select our three people, let's reorder them. Three people filling three positions can be ordered 6 possible ways. Diving the 210 combinations from above into the 6 non-distict orders leaves 35 DISTINCT possible combinations. for the democrats.Now back to the republican. Since there are 3 different republicans we can choose to join the team, we have a total of $35\times3=105$ possible combintations.

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