Question
The set of all permutations function composition".of any non-empty set A always forms a group under the operationTrueFalse
The set of all permutations function composition". of any non-empty set A always forms a group under the operation True False
Answers
True or false? For a set of $n$ distinct objects, the number of different combinations of these objects is more than the number of different permutations.
In a permutation. The order does matter. So this is a true statement.
Where this is talking about compliments and whether or not it's true or false. So we're talking about you have a set? A. So we'll call. We'll call this set a, but then, if there's anything outside of set A that's called a compliment, so it's it's gonna be completely outside of this black circle, so to speak, so the answer is true.
All right, So this is a true or false question when two sets intersect isn't always going to be a sub sect of that whole union, and the answer in this case is true.
So when we're thinking about true or false set questions, a good way to figure this out in your head if it doesn't quite make sense right away is to draw circles and see how it works. So let's do that with this question. So we have to say we have two sets A M B. And we are asked if the intersection of the two sets will always be a subset of the union, so a subset means it contains some over all of the elements in the set. So if 234 was a sets will, then three would be a subset because three is part of the sets, but it doesn't contain any elements that are not in the set, so 35 would not be a subset. So on the left, let's draw the union so the union will be everywhere they share, which in this case will be the entire circle, and the intersection will be on the right, right? This is a union B and a intersecting. The intersection is only going to be the middle where they share common ground, so we can notice that the intersection contains Onley elements that are in the union, right? They both contain the middle part of the circle, but it doesn't contain a single element that is not in the union. And that tells us that that's true, that the intersection of two sets is always going to be a subset of the union.
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