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For COMPETITION Number of Total Problems: 6. FOR PRINT ::: (Book)
Adam, Benin, Chiang, Deshawn, Esther, and Fiona have internet accounts. Some, but not all, of them are internet friends with each other, and none of them has an internet friend outside this group. Each of them has the same number of internet friends. In how many different ways can this happen?
Note that if is the number of friends each person has, then can be any integer from to , inclusive.
Also note that the cases of and are the same, since a map showing a solution for can correspond one-to-one with a map of a solution for by simply making every pair of friends non-friends and vice versa. The same can be said of configurations with when compared to configurations of . Thus, we have two cases to examine, and , and we count each of these combinations twice.
For , if everyone has exactly one friend, that means there must be pairs of friends, with no other interconnections. The first person has choices for a friend. There are people left. The next person has choices for a friend. There are two people left, and these remaining two must be friends. Thus, there are configurations with .
For , there are two possibilities. The group of can be split into two groups of , with each group creating a friendship triangle. The first person has ways to pick two friends from the other five, while the other three are forced together. Thus, there are triangular configurations.
However, the group can also form a friendship hexagon, with each person sitting on a vertex, and each side representing the two friends that person has. The first person may be seated anywhere on the hexagon Without Loss Of Generality. This person has choices for the two friends on the adjoining vertices. Each of the three remaining people can be seated "across" from one of the original three people, forming a different configuration. Thus, there are hexagonal configurations, and in total configurations for .
As stated before, has configurations, and has configurations. This gives a total of configurations, which is option .
Answer:
A student council must select a two-person welcoming committee and a three-person planning committee from among its members. There are exactly 10 ways to select a two-person team for the welcoming committee. It is possible for students to serve on both committees. In how many different ways can a three-person planning committee be selected?
Let the number of students on the council be . We know that there are ways to choose a two person team. This gives that , which has a positive integer solution of .
If there are people on the welcoming committee, then there are ways to choose a three-person committee.
A set of square blocks is arranged into a square. How many different combinations of blocks can be selected from that set so that no two are in the same row or column?
There are ways to choose the first square. The four remaining squares in its column and row, and the square you chose exclude nine squares from being chosen next time.
There are remaining blocks to be chosen for the second square. The three remaining spaces in its row and column and the square you chose must be excluded from being chosen next time.
Finally, the last square has remaining choices.
The number of ways to choose squares is but you must divide by since some sets are the same.
Two cubical dice each have removable numbers through . The twelve numbers on the two dice are removed, put into a bag, then drawn one at a time and randomly reattached to the faces of the cubes, one number to each face. The dice are then rolled and the numbers on the two top faces are added. What is the probability that the sum is ?
At the moment when the numbers are in the bag, imagine that each of them has a different color. Clearly the situation is symmetric at this moment. Hence after we draw them, attach them and throw the dice, the probability of getting some pair of colors is the same for any two colors.
There are ways how to pick two of the colors. We now have to count the ways where the two chosen numbers will have sum .
Sum can be obtained as , , or . Each number in the bag has two different colors, hence each of these three options corresponds to four pairs of colors.
Out of the pairs of colors we can get when throwing the dice, will give us the sum . Hence the probability that this will happen is .
Ignoring the numbers that do not affect the probability of the desired outcome (the ones that are not on top of the dice), say that the number on top of the first die is . For the sum of the numbers to be , the second die must have the number on top. There are remaining numbers that could be on top of the second die, of which are (since in all cases).
Thus, the probability of the sum of the numbers being is , so the answer is .
Eight points are chosen on a circle, and chords are drawn connecting every pair of points. No three chords intersect in a single point inside the circle. How many triangles with all three vertices in the interior of the circle are created?
To choose a chord, we know that two points must be chosen. This implies that for three chords to create a triangle and not intersect at a single point, six points need to be chosen. Therefore, the answer is which is equivalent to 28,
We first figure out how many triangles can be created. This is done by choosing lines out of the , which is equivalent to . However, some of these triangles have vertices on the circle. Therefore, the answer choice must be less than . The only one that is so is .