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@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.Damn you doing ECAPS 😭😭
A. Either all the elements in the subset leave same residue by 3 or they leave residues 0,1 &2. Now consider 3 sets: {1,4...298}, {2,5...299},{3,6...300}. For first subcase, if all residues are equal, we get 3*100C3 choices. For different residues you get 100^3 choices (from each set, one.).
B. Comes from A
C. Absolutely true (Because you have to choose 3 elements from 300)
D. (From A) False.
So the answer is ABC?
ya the ans is abc
firstly- by residues u mean remainder right, this took me a little while to get it
secondly- our aim is to form a subset where total sum of no.s is a multiple of 3. C1 is if all remainders are equal, by this u mean ki all no.s come from 1 grp toh aise 3 case banenge and har grp mei 100 no.s so 3*100C3 choices
third- this 100^3 is coming bc har ek set se ek no. to form the required set
wht about the case where we choose 2 no.s from 1 grp and 1no.s from the remaining 3rd grp ?
Yep (Sorry I should have elaborated.). When we choose 2 numbers from 1 group and one from another it will not be a multiple of three.
oh i see..hmm..
thnku , got it👍
+solved @SirLancelotDuLac
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