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1-a) In how many permutations of the 26 letters in the alphabet. do the patterns ads dog and sol not appear? b) In how many ofthe permutations of {1,2._ 10} exactly...

Question

1-a) In how many permutations of the 26 letters in the alphabet. do the patterns ads dog and sol not appear? b) In how many ofthe permutations of {1,2._ 10} exactly 3 of the even numbers and one of the odd numbers are in the correct position?

1-a) In how many permutations of the 26 letters in the alphabet. do the patterns ads dog and sol not appear? b) In how many ofthe permutations of {1,2._ 10} exactly 3 of the even numbers and one of the odd numbers are in the correct position?



Answers

Solve each counting problem. Make a list of all of the permutations of the letters $A, B, C, D,$ and $E$ taken three at a time. How many permutations should be in your list?

Were asked How many patient? 26 English alphabet. Mhm. The strings fish, right? Or birds? Other words. The strings. F I s H r a t or B I r d Find some terms here, snake. You will be the set of all patient of the 26 letters in the alphabet will define a one will be this set of all Brings painting. Yes, A two will be the set of all strings containing the word right. And a three will be the set of all strings containing the word bird. Kurtz. Let's find the magnitude of you do is the number of permutations of 26 letters. I mean letters. So we have that the magnitude of you This is going to be 26 factorial over 26 minus 26 factorial or more simply, just factorial are leaving. This one for now. Now finds the magnitude a one. Consider fish as one letter. After all, these four letters have to be next to each other in order for the string to contain the word fish. How many other letters are there? Well, we've used four. So there are 26 minus four or 22 another letter. So we're selecting 23 letters from the 23 letters. In other words, magnitude a one. This is 23 factorial over 23 minus 23 factorial gorgeous 23 factorial and again, I'll just leave it in this form. Find a magnitude eight to we use a similar procedure again, or treat rats as one letter. Then it follows that there are 26 minus three or 23 other letters. So we need to select 23 plus one or 24 letters, firmly 24 letters. So the magnitude eight to is 24 factorial over 24 minus 24 factorial or simply four factorial. And finally find the magnitude of the three. Consider Bird is one letter, and there are the word bird. Contains four letters settling. There's going to be 22 letters remaining. It's a total of 22 plus one or 23 letters. The magnitude of a three going to be 23 factorial over 23 minus three. Bacterial. Just simply 23 factorial. Alex. Consider other sets Seem to think about what about this at a one intersect, too. These are the set of all strings which contain both the word fish and the word rats. So well, consider the two words as one letter each. Then how many letters is this? Where we have f I s h, that score and our 80? It's three born. So this has money left over 26 minus seven or 19 letters. What? And therefore we want to select 19 plus two or 21 letters, probably 21 letters. So the magnitude a one intersect too is equal to 21 factorial now Likewise, we have the magnitude of a one intersect a three Well, we know that they want to respect a three contains the strings that contain both fish and bird. But think about this. If the string contains both fish and bird, then it follows that it has to contain at least two characters. I This is impossible because this is a permutation. Of the 26 letters, strings are and therefore a one intersect. A three is the empty set and it contributes zero a 268 3. These are the strings that contain both rats and birds. But again, we see then that it must contain at least two or good Rottenberg have an art. But this is impossible because this is a permutation extraneous of the 26 letters. So a 200 a three is also and also so it contributes zero. Clearly we have a one intersect A to intersect a three, obviously the whole set. Finally, we'll use the principle of inclusion and exclusion. So we have that we're looking for is the number of letters that do not contain any of the strings Fish, rabbit, bird, finest. We're going to take the total number of these strings and subtract from the number of permutations of the 26 letters that contain at least one of strings fish, rather bird. So this is the set a one union A to union A three. And the magnitude of this set is given by using the principle of inclusion and exclusion Magnitude of a one plus magnitude of a two plus magnitude. A three minus the magnitude they want intersect too, minus the magnitude they want intersect. A three minus the magnitude. A two intersecting three was magnitude a one intercept to intersect three. So we saw that the magnitude of a one This was 23 factorial magnitude of a two. This is 24 factorial, the magnitude of a three, it says. Also 20 degree factorial minus the magnitude of a one intersect a two we saw. This was 91 factorial and the rest of the terms R zero. So we'll leave this in this storm. But we saw earlier that total number of permutations Magnitude of you is 26 factorial. So our answer is going to be 26 factorial minus 23 factorial plus 24 factorial plus 23 factorial bias. It should be I don t want that and minus 21 exports. And if you plug this into a calculator, this is a huge number. This is 402 61 right? 359 72 336797 900800 000 I believe it's 402 Septal ian. 619. 6 billion 359.2 million. 782 Porcellian, 336 Brilliant. 797 billion 900 million 800,000. And these are the number of permutations. Not containing the strings bird rats or fish. Very, very large number

All right, so we have a B c d. So we're gonna choose five objects. So that's five factorial on top. We have one that repeats itself. So that's two factorial. And then we have 31 so they really don't. You can put a man if you want, but they're not gonna change anything, cause your small claim I want. So we're not gonna do that. So what does that mean? That means we have five times, four times, three times, two times one over to Tom's one, which those of course, will cancel out, and we're left with 60.

All right. So our list, this Tom has a B c three days and to ease so notice that we have a total of eight letters to pick from. So that means that eight factorial is on the top, and then we have some repetitions. We have three ds and to ease, so we're gonna have a three factorial on a two factorial down there so we can write that out so that you can see how that cancels out. Eight Factorial is on the top, and that just means eight. And then you descending down. Then we have three factorial into factorial and so you can cancel out. Well, three times two is six. So I can take the six away and then two and one I can take out the two in the one, and I could multiply together what is left. And so that's gonna be eight times. Seven times five times four times three for a total of 3000. 360

All right, so we have a B B. C. So we have six objects in this list, but there are some repetitions. And so we're gonna use our repetition formula. And that's gonna tell me that I'm going to do you six objects of six factorial on the top and on the bottom gonna do each separate one, so I'm gonna have three A's. So that's three factorial I'm gonna have to be. So that's two factorial and I have one. See? So that's one factorial, which is basically just warned. So I'm gonna write this out so we can see what this looks like. But we have six times, four times, three times, two times once. Oops. I skipped the bottom and there didn't know I don't know how to count today, six times, five times for comes three times. Do you come for on the bottom? We have three factorial. Three times, two times worn times, two factorial times, one factory. Now, if you look at that, you can start to cross out things that will cancel. For example, I see a three and we'll see a tube and I see a one. Of course, the ones don't matter now. Here I see a two and a four, so I can cross that out. And so really, I have six times, five times two, which is going to give me a total of 60.


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