$1$ can go to $A$ or $B$ or none. Similarly all. Answer $3^7-1$.

It can be solved like this. by fixing the cardinality of one set. we the sum becomes: $2^7+2^6.7+2^5.7C2.......+1-1=3^7-1$.

guys i don't get it. 1 can be in both A and B. I got 2^7*(2^7-1)

$4^7 - 2^7$ is what i get.

If $1$ is in both A and B, their intersection contains $1$ while the condition states that their intersection is exactly $\{5,7,8\}$.

can anyone tell what is really the solution of this problem?

Here is the easy generalization to this problem can be made by anyone , Find number of $k$ tuples $\{A_1,A_2,....A_k\}$ such that those are pairwise distinct and each of them are subset of $X$ where $X=\{1,2,3....n\}$ and there are $m$ elements in their intersection region. Solution: Total regions are $k+1$ where we can send at most $n-m$ numbers other than elements in intersection region. And as those m elements are not fixed so number of $k$ tuples is $f(k)=\binom{n}{m}[(k+1)^{n-m}+(-1)^{r+1}\sum_{r=2}^{k}f(k-r)\binom{k}{r}]$ with $f(0)=1$

UNORDERED pairs $\{A,B\}$ are required. Therefore the answer is $\dfrac{3^7-1}{2}$, since each pair is counted once as $\{A,B\}$ and once more as $\{B,A\}$.

I don't think its clearly mentioned whether its ordered or not.

Well, being denoted by a notation for sets and not tuples tend to mean the former. But it's just a factor of 2.

chaotic_iak wrote: But it's just a factor of 2. Which is the crux of the problem. The problem is SO VERY easy that minor mistakes (and mind you, missing out "just a factor of 2" is NOT MINOR) will also count much.

I'm sure an incomplete solution gets 6/7 points; heck, you can even say "if we are asked to find ordered pairs, then the answer is $3^7 - 1$; if we are asked to find unordered pairs, then the answer is $\dfrac{3^7 - 1}{2}$" and get full credit (or otherwise you're facing incompetent graders). After looking at source: ...yes, it's from a national olympiad, so it should be essay. The above remark should give you a perfect solution.

I got the answer 3^7 instead of 3^7 - 1 . How much marks do you guys think I will get out of 17 ?

12? It's clearly stated that $A \neq B$ in the problem.

harish_kappe wrote: I got the answer 3^7 instead of 3^7 - 1 . How much marks do you guys think I will get out of 17 ? ZERO

kauterry wrote: harish_kappe wrote: I got the answer 3^7 instead of 3^7 - 1 . How much marks do you guys think I will get out of 17 ? ZERO No i don't think you will get a zero. You still will get partial marks if have shown how you got $3^7$

Hey guys, I'm from karnataka, although some of the questions in our region were exactly similar to these but some questions were different, do you know where can i get solutions to problems which appeared in our region ?

You can get it here.

iarnab_kundu wrote: $1$ can go to $A$ or $B$ or none. Similarly all. Answer $3^7-1$. Hey arnab , I am unable to reason out why did you subtract 1 from $3^7$ . Please explain it to me. Thanks.

$A = B = \{5,7,8\}$ would be counted otherwise, hence why we need to exclude that case.

Thank you Ivan!