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How many seven-digit numbers can be formed with no repeated digits_How many seven-digit numbers with no repeated digits contain a 3 but notAn urn contains 20 red ba...

Question

How many seven-digit numbers can be formed with no repeated digits_How many seven-digit numbers with no repeated digits contain a 3 but notAn urn contains 20 red balls and 15 white balls. sample of five balls is selected, with none retured t0 the urn after selection.How many different samples are possible?Suppose all the red balls are numbered. How many samples contain 5 red balls?

How many seven-digit numbers can be formed with no repeated digits_ How many seven-digit numbers with no repeated digits contain a 3 but not An urn contains 20 red balls and 15 white balls. sample of five balls is selected, with none retured t0 the urn after selection. How many different samples are possible? Suppose all the red balls are numbered. How many samples contain 5 red balls?



Answers

How many odd numbers of at most three digits can be formed using the digits $0,1,2,3,4,$ and 5 without repetitions?

So we haven't earned. That contains 15 red And there are also 10 white And we're gonna select five. The order doesn't matter. So we're going to have combinations and we want to know what's The number of ways that we can draw out? five Red And there are a total of 15 And we want to select five. So we can either go through and do the combinations and the 15 factorial over the difference factorial divided by five factorial. I'm going to use my software and type in the 15. The combination Uh five. And just get the answer and we find out that that's equivalent to 3003. Now next question we want to get three red And we want to have to white which means there are 15 red to choose from and we want to choose three and multiply There are 10 white and we want to pick two groups to So we have our combination of 15, choose three and I'm just going to use the button to figure it out. That comes up to be 455. And instead of typing it again, I'm just going to store that as X. And then the 10 choose to I can use the combination but that's just going to be 10 times nine. And then that product divided by two and I hit to the power of so we've got to clear that out 10 times nine divided by two, which is going to end up in 45. So this is 45. And then we multiply that by the 455 and find out that there are a total of 20,475 ways to do that. Now finally we have The probability or not probability just the number of ways. And we're going to have to have at least four, which means we could have four red Or we could have five red. Now there's an implication here that we have to choose five. So we have to choose five. So if we're going to have four red, that means we have to have it one White. And this is what a lot of students will forget to do. So for red. We have that combination of their 15 reds and we want to choose four. And that means we'll also have Can choose one white. Plus we'll have the likelihood of choosing five red and we already know that that's 3,003 because we already did that. So 15, choose four. Go back up and let's just do that combination up here and 15 shoes four That comes out to be 1365. 10 choose one is just 10. And so when we combine all that together, that's a three. We have that 1030, times town, Plus We have the 3003. And that total comes out to be 16,000 653 and notice that this combination of these two events happening is less than this one event. There are more ways for this event to happen than these two events, some together to happen. Kind of interesting.

In this problem, we have to make find visit numbers sets that it is odd number, So which means that at the last place The possible number is one, three or five. The only possible number that we have to choose are from 12 34 five or six. So we have the number of the number of possibilities. Here is three. The number of possibilities here is now out of those remaining numbers, I can choose any number, So it is five here, it's four here, it's three here, it's two. So the total number of ways in which A 5-digit number can be made. Such center numbers orders two and 2, three and 2, four and 2 five and 2 three. So this comes us 120 Into three, which is 360.

So the caution is saying that we have to select three balls of each color that is red, white and blue. And there are six red balls. There are 55 balls and there are five blue balls. Right? So there are six red bulls, six red balls And there are five white balls And there are five blue balls. And from each color we have to select three balls, three balls We have to select and the number of ways of selecting three balls. Among the $663 among the five white, we have to select three white balls, that is five C three. And among the five blue ball, we have to select three Bus in 5 C3 days. So according to a fundamental principle, according 63 in 25 C three in 25 C three will be the number of ways of selecting the nine balls and three among the each color. And it will come out to be as 63 is 20 into 10 into 10. That is in 2000. Well, we have used this formula for NCR. Right, Thank you.

We have as described in the book, A container of white and red balls. There are 15 red and there are 10 white. So we're going to pulse and different ones out and determine how many possible ways we could do each pull. First off suppose we pull all red balls. That is five red balls. So if we've pulled five red balls and that means we had 15 choices and he chose five. So a combination 15 5 and out of all the white balls, we had 10 choices and shows. Zero. So by the fundamental principle of counting, also known as the multiplication principle in this book, we can multiply these two terms together to determine how many different ways we could pull five balls where all of them are red. And we would get 3003 different ways to do this. All right, that's the 1st 1 out of the way. Next up, suppose we pull three red balls, so that is 15. Choose three combination and we pull to white balls. That is, the rest of our set of five balls are white. Well, now we have a combination 15 3 and 10 to and we once again multiply these together to get a much larger number than before 20,475 different ways to do this. All right. Now it's time for the final one supposed that we pull at least four red balls. This is where things get a little bit interesting. So if there's at least four than it could be either four red balls or five red balls, suppose it's full red bulls. Then we have 15. Choose for for the red ones and 10 choose one for the white ones. That is, It's always going to be five balls we have one plus four equals five. However, this was at least four. So if it's at least four, then we also must account for the case where it's five, that is, 15 choose five and 10 to 0. So if we add these two up, then we're accounting for both cases. That is the case where it's four. Plus. The case where it's five will give us the total number of cases where it is, at least for and when added up, these two terms will give 16,653. Thus, there are 16,653 ways to choose at least four balls, and we are finished with this problem


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