FCP با محدودیت ها

فصل: بخش ریاضی / درس: اصل شمارش / درس 3

FCP با محدودیت ها

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FCP with Restrictions

The fundamental counting principle with restrictions. Many counting problems on the test contain restrictions of some kind. The test it’s not enough to say something like, put 5 children in order. The test loves to throw in some kind of restriction. This child must sit next to that child, or this child must sit in such and such seat, or so and so can’t be on the end.

Something like that. Counting problems with restrictions, that’s actually what you’re gonna be dealing with on the test. When a restriction of this sort appears in a counting problem that can be broken into stages. Always start with the most restrictive stage.

If more than one stage contains a restriction then start with the most restrictive, then the next most restrictive, et cetera until only unrestricted stages are left. Here’s a practice problem. Pause the video and then we’ll talk about this. Okay.

So we have seven children with nice alphabetical names. We’re going to sit them in adjacent seats. Dahlia must be in the middle chair and George must be next to Dahlia. So that’s our restriction. How many different orders can they be arranged? So notice that if there were no restrictions, then the order, the total number of orders would just be the product of seven times all the numbers less than seven.

Okay? That would be easy. But that’s not what we’re asked here. So we have to think about this. So first of all, Dahlia we’re gonna put in the middle. There’s only one choice for her. Now, George has to be next to Dahlia, but George could be on either side, so there’s two choices for George.

Well, once these two children are seated, then we can look at the unrestricted choices. The other five children are unrestricted, so we’ll just do Al next. How many places can Al sit? Well, there are 5 choices for Al. Once Al sits down, there are only 4 chairs left for Betty.

And so forth with each child there are fewer chairs, three for Chuck, two for Ed and then Fran would have to sit in the last chair. And so the fundamental counting principle tells us that we just multiply these numbers. Well 5 times 4 is 20 and 3 times 2 is 6. Then 2 times 20 is 40, 6 times 40 is 240.

And that’s the answer. So there are 240 different arrangements that satisfy this particular restriction. Here’s another practice problem. Pause the video and then we’ll talk about this. In a certain summer school program, there are five periods in the day.

Each student takes English, Math, History, Science, and Science Lab. In how many different orders can a student’s schedule be arranged given that science lab must immediately follow the science class. Okay, very interesting. So first of all, let’s think about these 5 periods. Where could the science class fall?

That science class, it could fall in the first, second, third, or fourth slot. But the science class could not fall in the fifth slot. Because then it would be impossible for there to be a lab after it. So, four slots are okay but the fifth slot is not possible for the science class. So for science there are four possibilities. And once we pick science then the science lab, we have no choice about that.

It has to come right after science. So now we pick those, those two, so those two slots are occupied and now the others are unrestricted. So for English we’d have 3 choices, then Math we’d have 2 choices, for History, we would have 1 choice. Fundamental counting principle tells us we multiply all these numbers together.

And of course this is just 4 times 6, which is 24. There are 24 possible schedules. Counting problems with restrictions. Now, we are starting to see problems that actually might appear on the test itself. So these two practice problems, conceivably, something very much like this could appear on the actual test.

In summary, when a counting problem contains restrictions in certain stages, always do the most restrictive stage first.

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