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?
1- 75 179 420 (Not sure about the last 2)
2- 24 48 96
3- (Got it wrong)
4- 13 10 12 (Easy one)
Does anyone remember the 1st question(any part) ? I’d like to try solve it again
Well last time they got the results in 10 days so I guess it will be the same this time as well.
Well last time they got the results in 10 days so I guess it will be the same this time as well.
btw this time zio-zco were conducted together and zio was also conducted online so we can expect the result much sooner than that
Weren’t they conducted online last time as well?
Well here’s how I did the third question.
I used the concept of recursion and contradiction… So lets just list the first three cases
1: 1
2: 21
3: 231, 321, 312
So for 4, lets use contradiction to get things done.
–.--.–.--
4! = 24 (Total no of ways)
So here are some of the combinations that aren’t allowed,
1 – -- – (As p=1 is not allowed)
2 1 – -- (As p=2 is violated) Also, notice why I haven’t written 1.2_._ as another option. This is so because it is already included in the previous case.
2 3 1 –
3 2 1 –
3 1 2 – (As in all these cases, p=3 is violated, and all other cases have already been included)
So a total of (the factorial is for different cases, example, 1234, 1324, 1243…) :
(1 * 3!) + (2 * 2!) + (3) = 11
hence the total no of allowed ways are 24 - 11 = 13
Now here’s where you see the magic, observe how in every case, the first set of known numbers are nothing but the allowed ways of the previous number!
If you didnt understand, observe the method again. Ill try to explain it again. We want to find the total no of ways, which is equal to total no of ways - wrong ones.
to find the total no of wrong ways, we start with total no of ways when p=1 itself is violated. Then we move to no of ways p=2 is violated, without recounting cases
p=1, and build on!
So, for 5,
1 * 4! = 24
1 * 3! = 6 +
3 * 2! = 6 +
13 * 1! = 13 +
Sum = 49
Hence total number of allowed ways are 5! - 49 = 120 - 49 = 71(yay!)
Building up for 6 and 7, we get 461 and 3447.
Hi, guys.
My answers were:
1 a. 63
1 b. 143
1 c. 319
2 a. 40
2 b. 52
2 c. 88
3 a. 71
3 b. 461
3 c. 3461
4 a. 12
4 b. 9
4 c. 12
aah finally someone who got 4.a as 12
Can someone explain the logic for question 2?