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Petr Puzanov2011-04-24 13:16:17
Yandex
Petr Puzanov, 2011-04-24 13:16:17

Yandex questions on career day st. petersburg?

Let's say stud. The group consists of N people. They need to calculate the average score of the group, while not one person from the group wants to tell their average score to anyone. How can a group calculate their average score in words (those without using pens and papers)?
A notebook was given for answering this question. I got it, but my solution is not the easiest. Who can offer one?
I'll post mine later!
There was another question: there are 100 people in line for a pack of felt-tip pens worth 50 rubles. 50 people have a banknote of 50 rubles. 50 people have a banknote of 100 rubles. Everyone needs to buy felt-tip pens. What is the probability that the seller can give change to all buyers?
Answer 100% did not pass - He can give change to everyone who needs it after everyone buys felt-tip pens, they don’t roll, tk. He must return the change immediately after the purchase.
What are your options?
I'm writing from my phone - errors are possible!

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5 answer(s)
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rPman, 2011-04-24
@rPman

the first task is solved in an elementary way :)
1. choose one leader (not necessarily among students)
2. The leader thinks of a random number in his mind (large enough, comparable to the usual average score * N) and says it to the first student 'in the ear'
3. Each subsequent the student adds his score to this number and reports the result to another
...
4. at the end, the last student also whispers his sum to the leader
5. the leader subtracts his number from the result and divides it by N
i.e. no one sees the whole picture, which means that they will not be able to carry out any calculations (even if someone accidentally hears someone's amount, this will not work).

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asm0dey, 2011-04-24
@asm0dey

Well, for the first task, if you take it literally, then it’s enough to show as many fingers as you have points))
Thus, the ball will not be NAMED

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mark_ablov, 2011-04-24
@mark_ablov

This is a classic, in fact, we need to make the correct set of brackets (opening - "+50rub", and closing - "-50rub").
And the number of options will be equal to the Catalan number for n = 50.
Numerically, this counts as 100!/(50!*51!).
These are successful cases, there will be - C(100, 50) = 100! / (50!)^2.
Answer - 1/51

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TLN, 2011-04-24
@TLN

A sheet of paper is taken and everyone takes turns writing down their scores in subjects. That is, if there are 5 people and 5 objects, then there will be 25 numbers on the sheet in random order. then we calculate the average score for them.

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burdakovd, 2011-04-24
@burdakovd

According to the second, there will be as many correct variants of the queue as there are correct bracket sequences, namely fib(50) (where fib(1) = 1, fib(2) = 2, fib(i) = fib(i-1) + fib (i-2)).
Proved by induction.
The total queue options are the number of combinations, 100! / (50!)^2.
The answer is fib(50) / 100! * (50!)^2
In numbers, it turns out something very small, like 2e-19

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