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A computer programming team has 43 members. Suppose seven Team members are Wamen and six arc men. How many groups of five can be chosen that contain three women and...

Question

A computer programming team has 43 members. Suppose seven Team members are Wamen and six arc men. How many groups of five can be chosen that contain three women and (WO men? marks)

A computer programming team has 43 members. Suppose seven Team members are Wamen and six arc men. How many groups of five can be chosen that contain three women and (WO men? marks)



Answers

Selecting a Committee There are 7 women and 5 men in a department. How many ways can a committee of 4 people be selected? How many ways can this committee be selected if there must be 2 men and 2 women on the committee? How many ways can this committee be selected if there must be at least 2 women on the committee?

Game helps seven woman's nine mins. We have to select them select groups of five, and we want at least one woman in the group. So how can you eat it up? Do you check that? We're gonna consider the group when there is only one woman informants When that is to women and three men, when these three women to men's when there is for women. One man when these five men in fine woman and no men's. So we calculate those numbers and that give us 4242. That's the amount of ways that we can arrange this for now. The next question as us How many do we need? Um, harmony groups can reform if that he sold so at least one minute. So the only case that it wouldn't count is this one. Because in this one, there is no man, and we want at least one. So we're gonna add up everything as before. But the last one tensile that US 4221

In this problem there eight women, 11 men, and we're going to find the number of five member committees. And when you see the word committee, basically it's a group of people where the order doesn't matter. So we're looking at combinations here. So the 1st 1 if it's all women, that means there are five women. So out of a total of eight women were choosing five. And then, if we put that in a calculator or computed using the combination formula, we end up with 56 for Part B. The committee has all men, so that would be five men out of a total of 11 men. So that's 11 C five, and that works out to be 462 for Part C. There are three women and two men. So for the three women part, we have eight C three, and for the two men we have 11 C two and we multiply these because of the fundamental counting principal. We have women and we have men, and when it's and you multiply the number of outcomes. So eight C three is 56 11. C two is 55 so we multiply those together and we get 3080 and then for the case in Part D. It's actually a lot more complicated, so I'm going to give myself some more room. So if you have no more than three men, then your possibilities are zero man or one man or too man or three men. So we need to add together all of these different possibilities for zero men. That means there are five women, and we already know that if there are five women there 56 ways that can happen. We found that in part a. If there's one man, that means there's one man and there's four women. So for the four women, we have a choose for and for the one man, we have 11. Choose one. If they're two men, that means there's also three women. So for the three women, we have a choose three, and for the two men we have 11 choose to and for three men, that means there's also two women. So for the two women, we have a choose to, and for the three men we have 11 to 3, so we have a lot of combinations to work out. So this ends up being 56 plus 70 times 11 plus 56 times 55 plus 28 times 1 65 and that adds up to 8526.

And this question Were you one day, gentlemen? Yeah. Mhm. R six and total women R. Seven and were given date number of men in commodity is occurred to Yeah. Three. And the number of woman. Yeah. In committee is equal to four. Certain people. Where is equal to Mhm. 6 60 multiplied by 74. Yeah. Did the answer this question? Thank you fortunate.

Okay. This question has is talking about different committees that could form where we take five members from this work population. So the first part says how many committees could reform of five if they're all men? So to do this, it should be noted that in a committee it doesn't matter who's in one spot. It just matters whose a part of the committee. So this is a combination problem. So we see here that we have nine men and we need to choose all five of them to be on the committee and nine shoes. Five is just equal to nine factorial over five factorial times for factorial, and that's equal toe 1 26 Then for Part B. It's the same thing, but now it's all women. So similarly, we have 11 women and we want to choose all five of them to be on the committee. So that's 11 factorial over five factorial times six Factorial, which is 4 62 then finally, Part C has a split it up between two women and three men. So for the men again, we have nine men, but this time we only want to choose three of them but also we have 11 women, anyone who choose two of them, so that's equal to 1 39 when we plug each of those combinations in.


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