How many different functions are there from a set with 10 elements to sets with the following number of elements?
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The cardinality of $B^A$ is the same if $A$ (resp. $B$) is replaced with a set containing the same number of elements as $A$ (resp. $B$). Show Set $b = |B$|. When $b \lt 2$ there is little that needs to be addressed, so we assume $b \ge 2$. Assume $|A| = n$. A well known result of elementary number theory states that if $a$ is a natural number and $0 \le a \lt b^n$ then it has one and only one base-$\text{b}$ representation, $$\tag 1 a = \sum_{k=0}^{n-1} x_k\, b^k \text{ with } 0 \le x_k \lt b$$ Associate to every $a$ in the initial integer interval $[0, b^n)$ the set of ordered pairs $$\tag 2 \{(k,x_k) \, | \, 0 \le k \lt n \text{ and the base-}b \text{ representation of } a \text{ is given by (1)}\}$$ This association is a
bijective enumeration of $[0, b^n)$ onto the set of all functions Since $[0, b^n)$ has $b^n$ elements, we know how to count all the functions from one finite set into another. Video TranscriptAlright for this problem we want to answer the question, how many different functions are there from a set with 10 elements to a set with blank elements? Where for a through D. We have a different number in the blank. So for part a we have to in the blank to answer this. It's rather simple. It's just going to be not 10 to the power of to it's going to be two to the power of 10. So Doing 2 to the power of 10, that's 1,024. So for part a there will be 1,024 different functions from a set with 10 elements to a set with two elements. For part B we have a three in the blank and all we're going to need to do is now we have three to the power yeah 3 to the power of 10 59,000 uh 49. Then for part C we have four going in the blank. So as you can probably guess Answer here is going to be 4 to the power of 10. It's going to be one million 48,000, 576 different functions. And for part D we will have five going in the blank, And the answer is going to be five to the power of 10 Or 9,765,000, 625 different functions. Question: Video Answer: Get the answer to your homework problem. Try Numerade free for 7 days  We don’t have your requested question, but here is a suggested video that might help. Related QuestionHow many different functions are there from a set with 10 elements to sets with the following numbers of elements? a) 2 b) 3 c) 4 d) 5
DiscussionYou must be signed in to discuss. Video TranscriptWe want to answer the question of how many different functions there are from a set with 10 elements to a set with blank elements. We have a different number in the blank. We need to answer this in the blank. It's simple. It is going to be two to the power of 10. That's 1,024 if you do two to the power of 10. There will be over one thousand different functions from a set with 10 elements to a set with two elements. We have a three in the blank and all we need to do is get three to the power. There are four going in the blank for part C. The answer is going to be 4 to the power of 10. It will be over one million functions. The answer will be five to the power of 10 or 9,765,000, 625 different functions. Let A be set with 10 elements. A. 2 Given that set contain only 2 elements. 10 elements in set A will map to 2 elements, which implies = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 1024 functions B. 3 Given that set contain only 3 elements. 10 elements in set A will map to 3 elements, which implies = 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 = 59049 functions C. 4 Given that set contain only 4 elements. 10 elements in set A will map to 4 elements, which implies = 4 x 4 x 4 x 4 x 4 x 4 x 4 x 4 x 4 x 4 = 1048576 functions D. 5 Given that set contain only 5 elements. 10 elements in set A will map to 5 elements, which implies = 5 x 5 x 5 x 5 x 5 x 5 x 5 x 5 x 5 x 5 = 9765625 functions How many different function are there with set with 10 elements?(d) 9,765,625 different functions.
How many oneThe total number is (84)⋅4! =1680.
How many different oneHow many functions are there from a 5-element set to a 7-element? this element, so the total number of possible assignments is 7 · 7 · 7 · 7 · 7=75 . Thus, (c) is the correct answer.
How many different functions are there from A to B?If a set A has m elements and set B has n elements, then the number of functions possible from A to B is nm. For example, if set A = {3, 4, 5}, B = {a, b}. If a set A has m elements and set B has n elements, then the number of onto functions from A to B = nm – nC1(n-1)m + nC2(n-2)m – nC3(n-3)m+….
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