## Task A-3.7.

We have ten white, and one red, blue, green, yellow, and purple card each. The white cards are indistinguishable from each other. On exactly one side of each card is the symbol $X$. In how many ways can we stack these cards one on top of the other so that no two cards are facing each other with the side that has the $X$?
## Solution.

First, we will arrange the cards without considering which side they are facing (for example, we can imagine that all are facing the side with the $X$ symbol downwards). Each arrangement can be constructed by placing the blue, red, green, yellow, and purple cards among the 10 white cards that we do not distinguish in order.

The blue card can be placed in 11 ways (before the first white card, after the first white card, after the second white card, ..., after the tenth white card).

The red card can be placed in 12 ways (now we have 11 cards and the red card can be placed before the first of them, after the second, ..., after the eleventh).

Similarly, the green card can be placed in 13 ways, then the yellow card in 14 ways, and finally the purple card in 15 ways.

Thus, the 15 cards can be arranged in $15 \cdot 14 \cdot 13 \cdot 12 \cdot 11$ ways.

Next, let's determine how many ways we can choose which side each card will face.

If the card that is on the bottom (i.e., the bottom card) is facing such that the side with the $X$ symbol is up, then all other cards must be facing such that the side with the $X$ symbol is up.

If the bottom card is facing such that the side with the $X$ symbol is down, then either all cards are facing such that the side with the $X$ symbol is down, or at least one card is facing such that the side with the $X$ symbol is up. In the latter case, assume that the bottom $k$ cards are facing such that the side with the $X$ symbol is down, and the $(k+1)$-th card is facing such that the side with the $X$ symbol is up. Then all cards above the $(k+1)$-th card must also be facing such that the side with the $X$ symbol is up.

The number $k$ can be any number from 1 to 14. Adding the two cases where all cards are facing such that the side with the $X$ symbol is up, or all cards are facing such that the side with the $X$ symbol is down, we get that the total number of ways to choose how the cards will be facing is 16.

Finally, the arrangement of the cards and the way they are facing are independent, so the total number of ways to arrange all the cards is $16 \cdot 15 \cdot 14 \cdot 13 \cdot 12 \cdot 11$.

Note: The solution has two parts. One part consists of counting the arrangements of the cards regardless of how they are facing, and the other part consists of counting the ways in which the cards can be facing. The student can write these parts in any order.

The first part of the solution can be argued in multiple ways. One way is to say that the 15 cards will occupy 15 positions and it is necessary to choose which position the blue card will be in, which can be chosen in 15 ways. After that, we choose which of the remaining 14 positions the red card will be in, and which of the 13 positions the green card will be in, etc. We conclude again that the number of arrangements is $15 \cdot 14 \cdot 13 \cdot 12 \cdot 11$.

Another way is to say that the number of arrangements of the cards (regardless of how they are facing) is equal to the number of permutations with repetition, where one type of object appears 10 times, and each of the other five types of objects appears once. According to the formula, the number of arrangements is $\frac{15!}{10!} = 15 \cdot 14 \cdot 13 \cdot 12 \cdot 11$.

## SCHOOL/CITY COMPETITION IN MATHEMATICS

4th grade - high school - A variant
January 5, 2015.

## IF THE STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE PROBLEM, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH IN AN APPROPRIATE MANNER.