#pnc

1 messages · Page 1 of 1 (latest)

dim lodge
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i was thinking f(a) can be 2,3,4 (3 options)
now from b c d e , any 3 elements are to be chosen to fulfil the onto condition and then the remaining letter can take any value
so the ans will become
3 * 4c3 * 3! * 4 = 288 but thats wrong

image.png
graceful narwhalBOT
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<@&1227988399579730072>

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Note for OP

+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.

tiny aurora
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Is it 180?

dim lodge
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indeed it is

tiny aurora
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Total onto nikle k wo wale minus kiye jisme 1 ara tha usme 2 case bnae

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Ek jisme bcde ko 2,3,4 dia

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dusre me bcde ki 1,2,3,4

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Or dono case me a ko to 1 dia hi

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Total me se ye 2 minus krke ara hai

dim lodge
dim lodge
dim lodge
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still not there

subtle glade
# dim lodge i was thinking f(a) can be 2,3,4 (3 options) now from b c d e , any 3 elements a...

Apart from a, baaki sabhko if you have to match the cases are:

  1. all 4 elements of co-domain are mapped and a has to map to one of the 3 choices which is 4!.3
  2. Remaining domain elements map up 3 elements leaving out one of the range elements which can't be 1; and then a maps onto that giving you 3 (numbers we can leave)*3(numbers jispe 2 values will be mapped) times 4C2 (the 2 values that will be mapped on the said element with two images) times 2! for the remaining
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Add 'em both up to get the answer

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24 times 3+3 times 3 times 6 times 2

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Which is 72+108 which is 180

dim lodge
subtle glade
fossil talon
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is a lot of casework ngl

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in any case 2 elements ko ek pe map karna hai

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either a is part of a pair or its akela

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but i think thats kinda long

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240-240/4 works

dim lodge
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it makes perfect sense to

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first we see the cases for a

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then we fulfil the onto condition

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and then the remaining element

fossil talon
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you're overcounting

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say f(a)=2

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and f(c)=1=f(d)

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toh you counted them twice na

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wait oops

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just look at this

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isme b will go to 4 obviously im too lazy to redraw it

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the links in blue are the triples you picked and mapped first

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and the one in red is the leftover element that wasn't mapped

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you consider these as unique cases but in reality they are the same

dim lodge
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but

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in case 1
step1 : a mapped to 2
step 2 : b to 1 ; c to 3 ; d to 4
step 3 : e to 3

in case 2
step1 : a mapped to 2
step 2 : b to 1 ; c to 3 ;e to 4
step 3 : d to 4
in case 2 (unlike case1) bce were chosen . here it is impossible for e to be mapped to 3 since i set the conditions like that . 4c3 * 3!

image.png
fossil talon
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wdym one-one lol

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doesn't it say onto in the problem

fossil talon
fossil talon
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i don't follow

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the two cases you drew are valid arrangements that would be a part of your total count but how does that address the overcounting lol

dim lodge
dim lodge
subtle glade
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Like where you went wrong :/

dim lodge
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nope 🥀

subtle glade
dim lodge
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and the one i attached in original msg

fossil talon
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i still don't get what you tried to do with the two images

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you're telling me that you made these two distinct functions

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mappings*

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but i'm telling you

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that each one of these mappings

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has a duplicate

subtle glade
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So...

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overcounting happens

dim lodge
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the whole purpose of 4c3 was to select 3 elements out of 4 to map further

dim lodge
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par kahan hai woh nahi pata

subtle glade
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is the point

dim lodge
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so u r saying b1 c3 e3

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thats not possible too

dim lodge
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finally

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@subtle glade got it

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thanxx

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+solved @subtle glade