How many onto functions are there from a set with 6 elements to a set with 4 elements?
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Number of functions is an important topic in sets. A relation in which each input has a particular output is called a function. If f is a function from set A to set B, then each element of A will be mapped with only one element in B. In this article, we come across the formula to find the number of functions from given sets and some solved examples. Show
Consider a set X having 6 elements and another set Y having 5 elements. Every element of set X will be mapped to one element in set Y. So each element of X has 5 elements to be chosen from. Hence, the total number of functions will be 5×5×5.. 6 times = 56. 1. Number of possible functions 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}. The total number of possible functions from A to B = 23 = 8 2. Number of Surjective Functions (Onto Functions) 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+….- nCn-1 (1)m. Note that this formula is used only if m is greater than or equal to n. For example, in the case of onto function from A to B, all the elements of B should be used. If A has m elements and B has 2 elements, then the number of onto functions is 2m-2. From a set A of m elements to a set B of 2 elements, the total number of functions is 2m. In these functions, 2 functions are not onto (If all elements are mapped to 1st element of B or all elements are mapped to 2nd element of B). So, the number of onto functions is 2m-2. 3. Number of Injective Functions (One to One) If set A has n elements and set B has m elements, m≥n, then the number of injective functions or one to one function is given by m!/(m-n)!. 4. Number of Bijective functions If there is bijection between two sets A and B, then both sets will have the same number of elements. If n(A) = n(B) = m, then number of bijective functions = m!. Solved Examples – Number Of FunctionsExample 1: The number of onto functions from set P = {a, b, c, d} to set Q = {u, v, w} is: (A) 68 (B) 36 (C) 81 (D) 64 Solution: P = {a, b, c, d} Q = {u, v, w} Here n(P) = m = 4 n(Q) = n = 3 The number of onto functions = 34 – 3C1(3-1)4 + 3C2(3-2)4 = 81 – 48 + 3 = 36. Hence, option B is the answer. Example 2: The number of bijective functions from set A to itself when A contains 106 elements is (A) 106 (B) 106! (C) 1062 (D) 2106 Solution: n(A) = m = 106 The number of bijective functions = m! = 106! Hence, option B is the answer. Related video Frequently Asked QuestionsLet set A has p elements and set B has q elements, then the number of functions possible from A to B is qp. If n(A) = n and n(B) = m, m≥n, then the number of injective functions or one to one functions is given by m!/(m-n)!. If n(A) = n(B) = p, then the number of bijective functions = p!. A function is one-to-one if every element of the range of the function corresponds to exactly one element of the domain of the
function. Solution: Functions are the backbone of advanced mathematics topics like calculus. Functions are of many types, like into and onto. Let's solve a problem regarding onto functions. To find the number of onto functions from set A (with m elements) and set B (with n elements), we have to consider two cases: ⇒ One in which m ≥ n: In this case, the number of onto functions from A to B is given by: → Number of onto functions = nm - nC1(n - 1)m + nC2(n - 2)m - ....... or as [summation from k = 0 to k = n of { (-1)k . nCk . (n - k)m }]. Let's solve an example. → Let m = 4 and n = 3; then using the above formula, we get 34 - 3C1(3 - 1)4 + 3C2(3 - 2)4 = 81 - 48 + 3 = 36. Hence, they have 36 onto functions. ⇒ One in which m < n: In this case, there are no onto functions from set A to set B, since all the elements will not be covered in the range function; but onto functions from set B to set A is possible in this case though. Hence, The formula to find the number of onto functions from set A with m elements to set B with n elements is nm - nC1(n - 1)m + nC2(n - 2)m - ....... or [summation from k = 0 to k = n of { (-1)k . nCk . (n - k)m }], when m ≥ n. Write the formula to find the number of onto functions from set A to set B.Summary: The formula to find the number of onto functions from set A with m elements to set B with n elements is nm - nC1(n - 1)m + nC2(n - 2)m - ... or [summation from k = 0 to k = n of { (-1)k . nCk . (n - k)m }], when m ≥ n. How many onto functions are there from a set with 6 elements to a set with three elements?Uh From six element to a set of three elements would be given by three days to six three. to raise to six mile plus three C 2 to raise two, sorry. Plus, It would be one race to 5, 1 days to six. So the final answer would be three days to 6 -3. 7.
How do you find how many onto functions there are?Number of Surjective Functions (Onto Functions)
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+…. - nCn-1 (1)m. Note that this formula is used only if m is greater than or equal to n.
How many onto functions are there from a set consisting of 4 elements to a set consisting of 2 elements?What are the number of onto functions from a set A containing m elements to a set B containing n elements. I found that if m=4 and n=2 the number of onto functions is 14.
How many onto functions from set A to set a?total no. of onto functions=n×n−1×n−2×2×1=n!
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