Suppose 51 cars start at a car race. in how many ways can the top 3 cars finish the race?

Suppose 51 cars start at a car race. In how many ways can the top 3 cars fnish the race? Tho number of different top three finishes possible for this race of 51 cars is (Use integers for any number in the expression )

Suppose 40 cars start at the Indianapolis $500 .$ In I how many ways can the top 3 cars finish the race?

In this question late and B the number of balls in you are in so it is equal to 39. Number of balls to select from this can be given by our, which is equal to five. So the number of which to select five balls from 39 balls randomly without replacement. Latest teeth given by and capital. And so it is given by 39 c five which so I can write devalue. The probability of winning a little low toe with one ticket is given by B is equal to one by 39 c. Fight solving it further I can write Evaluate one by 39 factories by five Factorial multiplication 39 minus five Factorial I just use the simple formula on solving it. I cannot evaluate one body 39 Multiplication 38 multiplication 37 multiplication 36 multiplication 35 multiplication 34 factorial by five Factorial multiplication 34 factorial. So this and this it can see loud. So I finally got the evaluate one by 575757 as our answer

23. We are interested in the number of different ways in which we select three people out of 50 where the order here is important. So we will use the permutation, um which is equal to 50 factorial over 50 minus three. Factorial, which is 50 factorial over 47 factorial, which is 15 times 49 times 48 times 47 factorial over 47 factorial. So this is equal 117600

In this question. We're being asked to figure out how many different finishes there could be for the first five places in a 50 person race. And it does say we're excluding ties, meaning we're assuming there is a single person for each place. Well, in order to figure out how many different ways there are to do this when we first have to determine is whether we're dealing with a combination or permutation, because certainly that changes what we're going to be doing. If we're doing a combination, that means the order does not matter if we're doing a permutation, that means the order does matter. Well, if we're thinking about this scenario, we're talking about the finishing of a race and what place you finish. In this scenario, I would say it's pretty definite that we are looking at something where the order does matter, right, because if you're running a race you want to get first place. First place is definitely better than second place, and you air of nothing else. You maybe wanna be top three instead of just being in the top five. But certainly anytime you're doing anything competitive, you don't want to just be one of the best people. The point is you're supposed to be trying to be the best. So certainly the order does matter here. Which means in this scenario, we're dealing with a permutation. Now that we know we're dealing with a permutation, all that's left is to figure out what our n and our values are going to be remembering that it's and the number of objects taken are at a time. Well, in this case, we have 50 people. But we are only interested in the top five, the positions that the first five people finishing so r n value is going to be 50 and our our value is going to be five. Meaning we're looking to do a permutation for 50 comma five do beware, depending on which calculator you are using will depend on how exactly you're plugging that in, because in some calculators you may be plugging it in mawr as 50. Then p for permutation than five. Just kind of depends which one of these you're writing based on which calculator you're using. But if you plug into your calculator correctly, you should get 200 and 54 million 200 and 51,000 200. That is how maney different finishes we have. Because if we do a permutation, a permutation of 50 choose five gives us this very large number. Which makes sense, though, because if you think about it, we have 50 different people. We're only interested in five. So that's a lot of different groups of five. And within those groups, we also have to consider all the different possible ways that those five could finish first to fifth. So that is gonna be a lot of ways, just makes sense.

So here we're doing, we have a choice between the finished shirts of a list of six automobiles, so it's going to be six is equal to six, three, Which is equal to six, Which is equal to 120.

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