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.

# ZIO 2016 Discussion

**excudeles**#2

Which questions were you able to solve? I attempted 2, 3 and 4 although I don’t remember the answers at all.

**equinox**#4

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

**vedant2080**#6

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

**excudeles**#10

Can anyone who solved the third question post their method? Looks like everyone’s getting different results on that one.

**dmhero**#11

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

**excudeles**#13

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.

**parthdhar**#15

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

**anukul**#16

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

**r4huln**#19

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.