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in Mathematical Logic by (37 points) | 28 views
$1$ . we will select ‘2’ from the set.
No given answer is 3

1 Answer

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I understood what they are trying to say.

Suppose we wrote $9$ numbers $1,2,3,4,5,6,7,8,9$ each on different piece of paper and then put all those papers in a box and we shuffle those pieces.

Now how many minimum number of pieces we need to take out from that box to ensure that the number written on any $2$ of the pieces that we are taking out from the box results in $even$ number without looking at the numbers that we are picking out.

here the name of the box is $S$. i.e. box=set.

The answer will $3$.


suppose we take any number from the box then it may be $even$ or $odd$.

if $even$ say $4$ then our job is done.

if $odd$ say $9$ then we need to pick another number from the box and so we go for Round 2.



In the pick for 2nd time

if $odd$ say $7$ comes then then our job is done. since $9+7 =16 = even$

if $even$ number comes say $4$ then we need to take out another number which can again be $even$ or $odd$. ($\because$ $4+9=13$ is odd). i.e. we need to go for Round 3.



in the 3rd pick

if an even number say $2$ comes then $2+4=6=even$

if an odd number say $5$ comes then $5+9=14=even$

We don’t need to go for round $4$


i.e. we need to take out atleast $3$ numbers from the box to ensure that any of those $2$ numbers result in $even$ sum.

hence $3$ is the answer.

by (4.1k points)
edited by
But they should have mentioned that.

We are picking up the numbers without looking at them.right?

And the question should be somewhat like this.I think.:)

What is the minimum number to be fully confident that we get an even number.
yes...correct. if we are able to look in the set then obviously the question would make no sense we could directly pick any $2$ numbers of our own choice.
Now a days, we have to not only answer the question but also need to make questions too :)
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