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An urn contains 4 red; 5 blue; and 6 yellow marbles. 3 marbles are selected without replacement: How many different ways can exactly 1 blue marble be selected?Quest...

Question

An urn contains 4 red; 5 blue; and 6 yellow marbles. 3 marbles are selected without replacement: How many different ways can exactly 1 blue marble be selected?Question 9Recall that there are 52 cards in a standard deck of cards and 4 of those cards are aces. How many different two card hands can be dealt that contain at least one ace?Question 101 pCalculate the number of different ten letter "words" that can be formed using all ten letters in the following word: SUBSTITUTE

An urn contains 4 red; 5 blue; and 6 yellow marbles. 3 marbles are selected without replacement: How many different ways can exactly 1 blue marble be selected? Question 9 Recall that there are 52 cards in a standard deck of cards and 4 of those cards are aces. How many different two card hands can be dealt that contain at least one ace? Question 10 1 p Calculate the number of different ten letter "words" that can be formed using all ten letters in the following word: SUBSTITUTE



Answers

$53-60$ Distinguishable Permutations distinguishable permutations.
Arrangements In how many ways can two blue marbles
and four red marbles be arranged in a row?

In this problem of basics of content delivery. We have to solve its problems involving combination and we have given problem five playing cards having the number 23 4, 5 and six. I'll suffer. And two cars are then drawn. How many different two cards hands are possible? Has he helped give up five Guards to 3? 456 They are well suffered and we have to don't do god's only. We have total choices five and out of five we have to choose only two guards. So using the combination we take two cards out of five. Using the farm lock C. N. C. R. And factorial upon and minus R. Factorial. Multiplying R. Factorial. So it will be five factorial upon 5. -23 factorial. Multi blank to victory, expanding this five factorial. Using the form law in factorial equal to And multiplying and -1 factorial. So it will be equal to five. multiplied by four. Multiplying three factorial upon three factorial, expanding this to victoria. We have to multiply by one. Now simplifying the expression we have a value equal to and this will be our final answer

Okay, So the number of ways that you could have five cards and a hand where three cards were from one suit and two cards or from another suit, you really have to break this down quite a bit. Those three cards from one suit those could be three hearts or three diamonds or three states or three clubs. And then when we get all those numbers in that whole part figured out, we're gonna multiply by because of the. And remember in counting when it's Andy multiplied by the never ways to get two cards from another suit. Well, if you've already taken three cards from one suit, you only have three more suits to choose from. So it could be three. Excuse me. Could be two cards from some suit that you haven't yet used. I don't know what it is or two cards from another suit you haven't yet used, whichever that is, or two cards from that other suit you haven't yet used. So there's only three options there because you already used once you, you only have three suits left. Now let's work the combinations in, so there are 13 hearts available. And if you're going to choose three. That would be 13. Choose three for or when we can't, we add, There are 13 diamonds available. You're choosing three. So this is 13. Choose three also, plus there are 13 spades. You're choosing three, so this is also 13. Choose three, and for clubs, it would also be 13 to 3. So that whole thing, added together, is going to be multiplied by the next part. So, uh, pick one of the suits that's left. So one of the suits that's left has 13 cards left in it, so that would be 13. Choose to. There's another suit that's left. It also has 13 cards left in it. So that would also be 13 shoes, too. And they're still yet one more soup that's left and there's still 13 cards in it. So that would be 13 shoes, too. So one or the other or the other. That's why we're adding okay, so if we simplify the left part, we have four times 13. Choose three, and then when we simplify the right part, we have three times 13 choose to, and when we find those numbers, we get four times to 86 times. Three time 78 Multiply all that together and you get 267,000 696.

Total cards given to us our fight on. We need today two cards at the time to make our to guard hand. Now we will use combination here. Therefore the number of hands the number off to gun hands possible. Uh, C five do that is take two from five and the formula for combination. Since you know, they're the formula for combination, see, and are taking our at the time from an elements is in factorial over r factorial into n minus r. Factorial. We have used this formula here and solving this formula. We get five factorial over to Victoria and I was here was two so two factorial into three factorial. Therefore, we can write five factorial as five into four into three factorial and to factory allies to into one which becomes too and three factorial as three factory. So three factorial is common to board a numerator and denominator. So we have Tortola, then possible vase before our two card hand

Okay, So if you're playing a card game and using a standard deck of cards, which has 52 cards, 13 in each of four suits and you're trying to find the number of ways you can get four cards from one suit and one card from another suit, you really have to break it down like this. You could get four hearts or four diamonds or four clubs or four spades and one other cart. I remember from the counting section in this chapter, if it's an or statement you're going to add, and if it's an and statement, you're going to multiply. So the number of ways we could get four hearts would be 13. Choose for because there are 13 hearts and we're choosing four of them. We're gonna add to that the number of ways to get four diamonds, which is also 13 shoes for because there are 13 diamonds and we're choosing four of them. Add to that the number of ways to get four clubs 13 choose for same idea and add the number of ways to get four spades 13 shoes for so we add all that together and then we multiply it by the number of ways to get one other card from a different suit. Now, if you have used one suit, then you can't use that suit anymore. So there are only 39 other cards in other suits, so this would be 39. Choose one if there are 13 cards in one suit than there are 39 cards that are not in that suit. Now we can simplify this a little bit, because when we had all these combinations together, we just have four of the same combination. So we have four times 13 shoes for times 39. Choose one, and we put those into our calculators and 13 choose four works out to be 7 15 39 choose. One works out to be 39 and when we multiply all three numbers together we get 111,540


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