## 9. U kinu

Nino, Miro, Bruno, Damjan, Ivana, and Ana bought movie tickets for the first six seats in the third row. Ivana and Ana want to sit next to each other. In how many ways can the six of them be arranged if they want to satisfy Ivana and Ana's request?

## Result: $\quad 240$
Solution, first method.

Let's answer the simplest question first: in how many ways can five children sit on five seats? Anyone of the five children can sit on the first seat, which is 5 ways, anyone of the remaining four can sit on the second seat, which is 4 ways, and so on.

Therefore, there would be a total of $5 \cdot 4 \cdot 3 \cdot 2 \cdot 1=120$ ways.

Now let's consider our problem: there are six children, but since Ivana and Ana want to sit next to each other, we can link them together, and then we can ask how many ways it is possible to seat these 5 blocks:

![](https://cdn.mathpix.com/cropped/2024_05_30_033072dbc63a78c728f4g-09.jpg?height=99&width=625&top_left_y=1087&top_left_x=384)

The answer would be: 120 ways. However, in each of the obtained ways, the girls can switch places, which results in 120 new ways, making a total of 240 ways.

## Solution, second method.

Ivana and Ana can sit on the 1st and 2nd seats in two different ways. Once they are seated, the remaining four friends can sit in $4 \cdot 3 \cdot 2 \cdot 1=24$ ways. Therefore, there are 48 ways in the case where Ivana and Ana are on the first two seats. If Ivana and Ana sit on the 2nd and 3rd seats, similarly, there are 48 ways. Since Ivana and Ana can sit on the 1st and 2nd, 2nd and 3rd, 3rd and 4th, 4th and 5th, or 5th and 6th seats, which is a total of 5 cases, there are $5 \cdot 48=240$ ways for all six friends to sit.