ZIO 2016 Discussion

general
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21

1- 75 179 420 (Not sure about the last 2)
2- 24 48 96
3- (Got it wrong)
4- 13 10 12 (Easy one)

22

i got the answer 40 for 2 a)
@sandy1899 can u plz tell how u solved 1a) ?

23

n any idea abt when the result would be declared?

24

Does anyone remember the 1st question(any part) ? I’d like to try solve it again

25

Well last time they got the results in 10 days so I guess it will be the same this time as well.

26

Well last time they got the results in 10 days so I guess it will be the same this time as well.

27

btw this time zio-zco were conducted together and zio was also conducted online so we can expect the result much sooner than that

28

Sweg level: ∞

2 Likes
29

Weren’t they conducted online last time as well?

30

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.

31

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

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aah finally someone who got 4.a as 12

33

Can someone explain the logic for question 2?

34

does anyone know where to find the question paper

35

is ans 4)13 10 12?

36
  1. 75
  2. 175
  3. 399 (I got 398 which is wrong)
  4. 40
  5. 52
  6. 98
  7. 68
  8. 461
  9. I dont know :stuck_out_tongue:
  10. 13
  11. 10
  12. 12 (ob)
    Guys I think these are the answers, please reply!
37

7.71
9.3447

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7.71 9.3447

39

does anyone know where to find the question paper???

40

Just posted my solution for the 3rd question here - http://stackoverflow.com/questions/33720871/find-the-number-of-5-digit-numbers-using-1-2-3-4-5-in-which-first-k-digits

What do you think the cutoff is going to be? I messed up on the first 2 questions.