pnc
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
28 Replies
@Apu
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.Is it 180?
indeed it is
Total onto nikle k wo wale minus kiye jisme 1 ara tha usme 2 case bnae
Ek jisme bcde ko 2,3,4 dia
dusre me bcde ki 1,2,3,4
Or dono case me a ko to 1 dia hi
Total me se ye 2 minus krke ara hai
total = 5c4 * 4! * 4
4c3 * 3! * 3
4!
still not there
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
Add 'em both up to get the answer
24 times 3+3 times 3 times 6 times 2
Which is 72+108 which is 180
got step1
dk what u mean by "remaining domain elements" in step 2
Matlab basically a ko chorke Baaki sabh elements in set A
man that
is a lot of casework ngl
in any case 2 elements ko ek pe map karna hai
either a is part of a pair or its akela
but i think thats kinda long
240-240/4 works
i still dont get why my approach is wrong 🥀
it makes perfect sense to
first we see the cases for a
then we fulfil the onto condition
and then the remaining element
bhai i told you uss din bhi
you're overcounting
say f(a)=2
and f(c)=1=f(d)
toh you counted them twice na
wait oops
nvm
just look at this
isme b will go to 4 obviously im too lazy to redraw it
the links in blue are the triples you picked and mapped first
and the one in red is the leftover element that wasn't mapped
you consider these as unique cases but in reality they are the same
this isnt one-one
but
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!
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!
wdym one-one lol
doesn't it say onto in the problem
oh you didn't see this shayad
why is it impossible for e to be mapped to 3 though
i don't follow
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
whi yaar
because of the 3!
i basically wanna imply distributing 3 balls ( here 1,3,4) between 3 out of chosen boxes ( here bce)
b to 1 , c to 3 , e to 3 would imply one ball going into box c and box e at the same time
Did you get this?
Like where you went wrong :/
nope 🥀
Can you write out your approach once again?
this is the best i could
and the one i attached in original msg
i still don't get what you tried to do with the two images
you're telling me that you made these two distinct functions
mappings*
but i'm telling you
that each one of these mappings
has a duplicate
Case 1 mein imagine in step 2 it is e that's mapped to 3 and c is mapped to 3 in step 3
So...
overcounting happens
arey step 2 mein e kaise map ho jayega jab select hi nahi kia
the whole purpose of 4c3 was to select 3 elements out of 4 to map further
obv answer is not matching so overcounting hai
par kahan hai woh nahi pata
But unn 3 mein hi agar e choose ho jaye instead of c
is the point
so u r saying b1 c3 e3
thats not possible too
OHHHHHHH
finally
@Lance who? got it
thanxx
+solved @Lance who?
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