4 votes 4 votes Let f : A → B be function, where A = {1,2,3,4,5,6} and B = {1,2,3,4,5}. If f(1) = 4 then how many surjective (onto) functions are possible ? Set Theory & Algebra zeal set-theory&algebra functions zeal2019 + – Prince Sindhiya asked Nov 11, 2018 • edited Mar 6, 2019 by ajaysoni1924 Prince Sindhiya 1.2k views answer comment Share Follow See all 7 Comments See all 7 7 Comments reply Show 4 previous comments Magma commented Nov 27, 2018 reply Follow Share can you tell me that how to subscript and superscript ? I didn't find that func button in this updated version 0 votes 0 votes Shaik Masthan commented Nov 27, 2018 reply Follow Share check this 1 votes 1 votes Magma commented Nov 27, 2018 reply Follow Share Ohk fine :p earlier it was bit easy because both function are very frequently use so sir make a separate button for this two func btw thanks 0 votes 0 votes Please log in or register to add a comment.
Best answer 6 votes 6 votes related theory https://math.stackexchange.com/questions/334420/number-of-onto-functions formula derivation is in given link Gurdeep Saini answered Nov 11, 2018 • selected Nov 11, 2018 by Prince Sindhiya Gurdeep Saini comment Share Follow See all 18 Comments See all 18 18 Comments reply Prince Sindhiya commented Nov 11, 2018 reply Follow Share @gurdeep same solution is given for it But i understood the second part that f(1) =4 that's why we have excluded that but why we are considering 4 in first part and why this 5Factorial is coming i wrote it's solution as 240 only please tell me the first part by explaining it a bit more 0 votes 0 votes Gurdeep Saini commented Nov 11, 2018 reply Follow Share Prince Sindhiya F(1)=4 is given and set b contain 5 element now in set A we have total 6 element including 1 means 5 element excuding 1 given that 5 element in B it means both set have 5 ,5 element where 1 is already related to 4 now in the remaining 5 element of A we have to relate every element of A only one element of B here we can not relate 2 element of A to the single element of B because it we do so than this will not satisfy onto property because some element will left unrelated in B so first element of A can relate to 5 ways and second element of A can relate to 4 ways 3 rd element 3 ways 4th element 2 ways 5th element 1 ways so total ways is 5! so in this process (one thing you should note in this process 4 element of B is related to two different element of A ) 2 votes 2 votes Prince Sindhiya commented Nov 11, 2018 reply Follow Share Gurdeep thnxx for such a beautifull solution 1 votes 1 votes Magma commented Nov 11, 2018 reply Follow Share Gurdeep Saini tell me one thing we exclude (1) from A is understandable but why you exclude (4) from B ?? for every element y in the codomain B of f there is at least one element x in the domain A one of the element from A {2,3,4,5,6} can map to to B (4) right ?? 0 votes 0 votes Magma commented Nov 11, 2018 reply Follow Share Prince Sindhiya clear my doubt 0 votes 0 votes Gurdeep Saini commented Nov 12, 2018 reply Follow Share Magma element 1 of A already map to 4 of B now we can ignore that elment and do the maping of rest 0 votes 0 votes Magma commented Nov 12, 2018 reply Follow Share we can ignore 1 but why we ignore 4 also ?? because if 4 have 2 image in domain..then there's no problem right ??? I know I'm wrong....but I want clarify my doubt..that's why I asking question 0 votes 0 votes Gurdeep Saini commented Nov 12, 2018 reply Follow Share because if 4 have 2 image in domain..then there's no problem right ??? @magma this in included in first case 0 votes 0 votes Magma commented Nov 12, 2018 reply Follow Share what I wanted to say is that this is possible right ?? 0 votes 0 votes Gurdeep Saini commented Nov 12, 2018 reply Follow Share yes @magma this is possible but included in first case in the first case 1 element and one other of A related to 4 of B (shown by you in figure) but in second case only 1 element of A relate to the 4 element of B and rest elment of A relate to the remaining four element of B 0 votes 0 votes Magma commented Nov 12, 2018 reply Follow Share thanks 0 votes 0 votes Shadan Karim commented Nov 13, 2018 reply Follow Share @Gurdeep Saini , great solution , so in this we have to take 2 cases, one with element 4 in the set B and one without element 4 in set B , because f(1)=4 , so element 4 from set B could be used or could not be used, so 2 cases. 0 votes 0 votes Gurdeep Saini commented Nov 13, 2018 reply Follow Share @shadan yes 0 votes 0 votes Shaik Masthan commented Nov 27, 2018 reply Follow Share @Gurdeep Saini Good one :) 0 votes 0 votes Gurdeep Saini commented Nov 27, 2018 reply Follow Share i request to the person who downvoted this answer please tell me what is wrong so that i dont repeat this mistake in exam 0 votes 0 votes Shaik Masthan commented Nov 27, 2018 reply Follow Share @Gurdeep Saini who just down vote but didn't participate in the discussion, doesn't get this comment as notification 0 votes 0 votes Gurdeep Saini commented Nov 27, 2018 reply Follow Share @Shaik Masthan he/she should give the reason what is wrong in this answer 0 votes 0 votes Shaik Masthan commented Nov 27, 2018 reply Follow Share we can't do anything... it's his/her responsibility.... forget about it.. there is No problem with this solution 1 votes 1 votes Please log in or register to add a comment.