The first example of combinations is this.

## Remember our formula of combinations

C(n,r)=\left( \begin{array}{c}n\\r \end{array} \right) = \cfrac{n!}{r!\left(n - r\right)!}

Let’s start with the paragraph of our second example of combinations:

*From an urn, with 6 numbered white balls of 1 up to 6 and 7 black balls from 1 up to 7, at the same time, 4 are taken out. What is the probability of obtaining:*

*Balls with numbers whose product is even?*

To ensure that the product is even, the following must be fulfilled as shown in the following table:

\begin{array}{| c | c | c | c |} \hline \text{PAIR} & \text{PAIR} & \text{PAIR} & \text{PAIR} \\ \hline \text{PAIR} & \text{PAIR} & \text{PAIR} & \text{ODD} \\ \hline \text{PAIR} & \text{PAIR} & \text{ODD} & \text{ODD} \\ \hline \text{PAIR} & \text{ODD} & \text{ODD} & \text{ODD} \\ \hline \end{array}

Let’s think a little bit to know if the result will be even.

- If we multiply 4 even numbers, the result will always be an even number.
- If we multiply 3 even numbers and 1 odd number, the result will always be an even number.
- If we multiply 2 even and odd numbers, the result will always be an even number.
- Finally, if we multiply 1 even number and 3 odd numbers, the result will always be an even number.

If you need to make a table to better visualize the balls, draw a table. We made the following table:

\begin{array}{|c|c|} \hline \text{White balls} & \text{Black balls}\\ \hline 1 \ 2 \ 3 \ 4 \ 5 \ 6 & 1 \ 2 \ 3 \ 4 \ 5 \ 6 \ 7\\ \hline \end{array}

And as we have only 4 ways to make the product result in an even number, we have to put the 4 cases in the following way with the operations already done:

\begin{array}{|c|c|c|} \hline \text{Case number} &\text{Case with letters} & \text{Caso with numbers}\\ \hline \hline 1 &\text{All pairs} & 6C4= 15 \\ \hline 2 &\text{Three pairs and one odd} & 6C3*7C1= 140\\ \hline 3 &\text{Two pairs and two odd} & 6C2* 7C2 = 315\\ \hline 4 &\text{A pair and three odd} & 6C1*7C3 = 210\\ \hline \end{array}

Consider that of the total of 13 balls, we have 6 with even number and 7 with odd number.

As in case 1 all the numbers of the balls have to be even and in total we only have 6 balls that have even numbers, the operation was made 6\text{C}4 that gives us as result 15 combinations.

In case 2 you need 3 pairs and 1 odd, then what you have to do is combinations of 3 even balls and combinations of 1 odd ball, in mathematical terms is 6\text{C}3*7\text{C}1 which results in 140.

In case 3 you need 2 pairs and 2 odd, so you can easily apply combinations of 2 even balls and combinations of 2 odd balls, you would mathematically do 6\text{C}2*7\text{C}2 which results in 315.

In the last case you have to leave 1 ball with even number and 3 with odd number, we simply apply 6\text{C}1*7\text{C}3 which results in 210.

We have to sum all the results that we just obtained, to have a total of 680. Then what you have to do is apply combinations to know all the possible cases that there are to get 4 balls from the 13 there is, so applying 13\text{C}4 results in 715 combinations.

Finally what you have to do is divide the sum of all the combinations calculated between the number of combinations that can be obtained from 4 balls of the 13:

\cfrac{680}{715} = 0.951

So the probability of getting 4 balls whose product is even is 0.951

*Balls with numbers less than 2?*

The answer is the following:

In this case the probability is zero because we only have two balls that are smaller than 2, that is, the balls with the number 1.

*Balls with numbers greater than 2?*

This paragraph is quite simple because all you have to do is discard the 4 balls that have numbers less than or equal to 2. Instead of having 13 balls, now we will have 9 balls and apply the following combination:

9\text{C}4=126

And the result obtained has to be divided between 715 which is the total of combinations that there are to grab 4 balls of the 13:

\cfrac{126}{715}=0.1762

So the probability of taking balls with numbers greater than 2 is 0.1762

*Balls with different numbers?*

Of all the balls, we must bear in mind that there are 6 repeated numbers, so it is necessary to know the number of possible combinations that can be made by excluding the ball with the 7 number:

6\text{C}4 = 15

The number of combinations that can be alternated with white and black balls is calculated:

(2\text{C}1)^{4}

The calculation that was made was because we have 6 repeated balls, that means that for the number 1 we have 2 combinations: that is white or black. For number 2 we have 2 combinations: that is white or black. And so with all the others. Why is the exponent 4 and not 6? Because we’re only going to extract 4 balls, not 6.

Then proceed to multiply the number of possible combinations that can be done by excluding the ball with the number 7 with the number of combinations that can be alternated with the white and black balls:

6C4*(2C1)^{4}=240

Now consider the 7 that had been discarded. We identify he number of combinations that can be made with 6 pairs and the 7:

6\text{C}3

We wrote a 3 because a place of those 4 we have already occupied because we are including 7.

Then calculate the number of combinations that can be alternated with white and black balls, of the 4 possible, 3 are repeated and 1 is not:

6\text{C}3*1\text{C}1*(2\text{C}1)^{3}=160

Of the two events calculated (when it is considered and when it is not considered the number 7), we have to sum them:

6\text{C}4*(2\text{C}1)^{4}+6\text{C}3*1\text{C}1*(2\text{C}1)^{3}=400

The 400 obtained has to be divided among the total of combinations that there are to remove 4 balls from the 13 (we already calculated it, remember that 13\text{C}4 equals 715):

\cfrac{400}{715}=0.5594

So the probability that different balls can be extracted will be 0.5594

**Thank you for being at this moment with us:)**