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Exercise of combinations

Let’s start with the subsection of this exercise of combinations:

A university wants to organize a series of conferences for 15 people in your favorite state. The problem is that the hotel is saturated, so they only got three rooms of different sizes. In the presidential suite they decided to accommodate 6 participants, in a junior to 5 and in a double room to another 4. The Committee wants to know how many ways it is viable to accommodate the attendees.

For the realization of this exercise of combinations,you have to make 3 combinations. The first combination is with respect to the room of the suite, the second is with respect to the room of the junior and the third is with respect to the double room.

Remembering the formula of combinations:

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

We will represent all the calculations mentioned in a table:

\begin{array}{| c | c | c |} \hline\\ \text{Suite} & \text{Junior} & \text{Doble} \\\\ \hline \\ \text{C}(15,6) = \left( \begin{array}{c}15\\6 \end{array}\right) & \text{C}(9,5) = \left( \begin{array}{c}9\\5 \end{array}\right) & \text{C}(4,4) = \left( \begin{array}{c} 4 \\ 4 \end{array}\right)\\\\ \hline \\ \ \left(\cfrac{15\cdot 14 \cdot 13 \cdot 12 \cdot 11 \cdot 10}{6!} \right) \ & \ \left(\cfrac{9\cdot 8 \cdot 7 \cdot 6 \cdot 5}{5!} \right) \ & \ \left( \cfrac{4\cdot 3\cdot 2 \cdot 1}{4!} \right) \ \\\\ = 5005 & = 126 & = 1 \\\\ \hline \end{array}

Let’s explain the calculations made in the table above. As we started to accommodate people in the Suite, that means we have 15 people to accommodate but there are only 6 spaces, so combinations were applied to know how many ways can accommodate 15 people in 6 spaces.

Once we accommodate the 6 people in the suite, there will be 9 people to accommodate in any of the other two rooms and as the next room we chose is the junior, then combinations were applied to know how many ways can accommodate 9 people in 5 spaces.

Finally, after accommodating those 5 people, we will have 4 people to accommodate in the double room, so combinations were applied to know how many ways can accommodate 4 people in 4 spaces.

The values obtained must be multiplied to obtain the total of viable ways to accommodate the attendees:

(5005)(126)(1)= 630630

Thus we obtain that we have 630630 viable ways to accommodate the attendees in the three rooms.

It does not matter if you start to fit in with the junior or the double, the result will always be the same. Check it out!

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