How was the exam for all of you ?
I don’t think that I’ll get selected, it was a easy paper but wasted alot of time.
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Which questions were you able to solve? I attempted 2, 3 and 4 although I don’t remember the answers at all.
Me too, i made a stupid decision to not try question 4 alot.
It was bad for me too… 
So guys what are the answers ya’ll got?
- Binary : 80 , 192 , 448 == All wrong… Over counted repetitions by mistake
- RGB Board game : 42 , 58 , 76 == No idea how…
- King and Minister : 71 , ~400(forgot :P) , 3447 == Used recursion… need someone to verify though
- Lists : 13 , 10 , 12(NOT SURE OF ORDER) == Easy one
Guys please check and share your results… What do you think the cut off is going to be this year?
I got these answers. Maybe some are wrong.
1.a 75 1.b 175 1.c 398
2.a 44 2.b 58 2.c Did not attempt
3.a 80 3.b 512(most probably) 3.c Forgot what i got. It was a 4 digit number starting with 3 and ending with 8. 
4.a 13 4.b 10 4.c 12
My answers are-
1- a-80 b-192 c-448 (All wrong)
2- a-40 b-52 c-96 (Not sure whether correct or not)
3- Did not attempt
4- a-13 b-10 c-12 (Seems to be correct after seeing other’s answer)
Please somebody tell me how much did i scored and what are the correct answers
can any1 remember what was the binary subset in the 1st question?
then the answers are 1 a) 80
b)192
c=448
n hey what are ur respective classes?
Can anyone who solved the third question post their method? Looks like everyone’s getting different results on that one.
well i did them like that…
let X=11011
so a binary sequence of 9 dig would be of forms
X _ _ _ _
_ X _ _ _
_ _ X _ _
_ _ _ X _
_ _ _ _ X
so this is a case of permutations where repitition is allowed
so the permutations of 0 and 1 in the blanks would be 2^4
as the position of x is independent of position of the permutations of the other binary digits,
the ans is 5*(2^4)=80
can someone post the questions?
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I believe your method is wrong. You’re doing some double counting this way. For instance, (11011)0110 is the same as 110(11011)0.
Today’s P3 (King and Minister) was actually INMO 1992 P4 (k=6).
You can check out the discussion on AOPS here.
These are called indecomposable permutations; one can solve for the number of them of a given length using generating functions, or use a recursion. They are given by sequence A003319 in the OEIS.
The answers were:
- 71
- 461
- 3447
1 a. 80 b. 192 c. 448 - ✗
2 a. 40 b. 52 c. 96 - ✓
3 a. 71 b. 461 c. 3447 - ✓
4 a. 13 b. 10 c. 12 - ✓
I think I scored 60, I’m in class 12.
Also, this
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Guys what about problem 2?
Great to know I know I got P3 fully correct 
What do you reckon the cutoff will be?
I think in the range of 25-35 from 10th to 12th so 25 for 10th and 30 for 11th and 35 for 12th as last year’s was much easier and the cutoff range was 35-45.
when r the results supposed to come out?