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The problem is as follows:

In a jar there are tickets of the same size and color which have a number printed from $10$ to $\left(4n+10\right),\, n\geq 2,\,n \in > \mathbb{N}$ .How many tickets could be taken out at random from the jar the least possible to be certain that among the tickets extracted there are $3$ having an odd number printed?.

The alternatives given in my book are as follows:

$\begin{array}{ll} 1.&2n+1\\ 2.&2n+2\\ 3.&2n+3\\ 4.&2n+4\\ \end{array}$

I'm confused about this problem. What I believe the procedure to solve this is to think about the worst case and add to that the elements requested so we can claim that we have what it is being asked.

This task could be much easier if I had known the number of elements explicitly let's say a number belonging to $n \in \mathbb{N}$, but having it defined by a constant not indicated by a formula.

The next thing which I attempted to do was that the worst case would be taking out all even numbers from $10$ to $\left(4n+10\right),\, n\geq 2,\,n \in \mathbb{N}$ and adding to that $3$ as then we can claim we have three odds.

But since the number of elements is not given explicitly I could obtain them by subtracting the total from the number of all odds and adding $3$ to that.

Thus I counted (considering not making the fencepost error).

Total:

$\left(4n+10\right)-10+1=4n+1$

Total of all odds:

$11+(n-1)2-11+1=2n-1$

Therefore the total needed to be extrated from that jar would be:

$4n+1-(2n-1)+3= 2n+5$

But this answer doesn't appear within any of the alternatives. What could be the part where I got it wrong?. Can somebody help me with this?.

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There are $2n$ odd numbers and $2n+1$ even numbers.

Therefore $2n+3$ numbers could be chosen and you might still have found only two odds.

So $2n+4$ are required.

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Out of $4n+1$ numbers, $2n+1$ of them are even and $2n$ of them are odd. Note that the first number is even, there are more even numbers than odd numbers. Also the number of odd numbers and even numbers can only differ by at most $1$.

$$(4n+1)-(2n\color{red}{-0})+3=2n+4$$

Remark: When $n$ is not given and if you are more comfortable with explicit numbers, you can choose your own $n$ to get started.

Edit:

Let's use another notation $N$ (another notation is needed as $n$ has already been used) to denote the number of odd numbers.

Using the formula of $AP$, and knowing that hte first odd number is $11$ and $4n+9$.

$$11+2(N-1)=4n+9$$

would give us that $N=2n$.

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  • $\begingroup$ How do I arrive to the conclusion that $2n+1$ are even? What exactly went wrong with my statement? I mean I assumed that I could get the all odds by subtracting the total minus the evens. Could it be that should I had ignored this and just use the a general formula that all odds are $2n+1$ and evens $2n$ always?. $\endgroup$ – Chris Steinbeck Bell Nov 7 '19 at 8:44
  • $\begingroup$ $4n+1$ divided by $2$ is $2n+0.5$. So you either have $2n$ or $2n+1$ even numbers. Since the first number is even, the number of even numbers increment faster than the odd numbers as you count. Hence you have $2n+1$ even numbers. $\endgroup$ – Siong Thye Goh Nov 7 '19 at 8:47
  • $\begingroup$ I see, but did my reasoning was incorrect? or was just the application on this particular problem would not yield a solution?.The second part about increments is where I'm stuck at. How can I tell which is faster if I don't know the end beforehand?. I mean the last number is $4n+10$ or do I arrive to that conclusion because that number will necessarily be even, hence I started with an even and end with also an even, can I claim there will be less odds?. $\endgroup$ – Chris Steinbeck Bell Nov 7 '19 at 9:01
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    $\begingroup$ I do not know how do you count the number of odd numbers. Let $e$ denote even and $o$ denote odd. If it starts from $e$ and ends with $e$, you will have more even numbers, suppose you end with odd numbers, the numbers are equal. Try to pair the even and odd numbers up. $(e,o,e,o,e)$ we have $(e,o), (e,o), e$, there is one unpaired even number, the additional even number. $\endgroup$ – Siong Thye Goh Nov 7 '19 at 9:06
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    $\begingroup$ I see, if you want to use the AP formula, let's solve for $N$ in $11+(N-1)2=4n+9$ which reduces to $2(N-1)=4n-2$, and hence $N=2n$, there are $2n$ odd numbers. $\endgroup$ – Siong Thye Goh Nov 7 '19 at 9:29

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