High Fives

a. Five “mathletes” celebrate after solving a particularly challenging problem during competition. If each mathlete high fives each other mathlete exactly once, what is the total number of high fives?

b. If n mathletes shake hands with each other exactly once, what is the total number of handshakes?

c. How many different ways can five mathletes be seated at a round table? (Assume that if everyone moves to the right, the seating arrangement is the same.)

d. How many different ways can n mathletes be seated at a round table?

Five mathletes are celebrating together.

a.

The combination rule helps to compute the number of possibilities of selecting r units from a total of n units. In this case, units are selected without replacement. The order of the arrangement of the selected r units does not matter.

The combination rule is applied using the given formula:

${}^n{C}_r=\frac{n!}{\left(n-r\right)!r!}$

Here, the total number of mathletes is five.

The number of mathletes required for a high five is two.

Since the mathletes high five each other only once and the order does not matter, the rule of combination is applied as shown below:

$$\begin{gathered} {}^5{C}_2=\frac{5!}{\left(5-2\right)!2!} \\ =10 \end{gathered}$$

Therefore, the number of high fives possible is equal to 10.

b.

The total number of mathletes is n.

The number of mathletes required for a handshake is two.

Since the mathletes shake hands with each other only once and the order does not matter, the rule of combination is applied as shown below:

$$\begin{gathered} {}^n{C}_2=\frac{n!}{\left(n-2\right)!2!} \\ =\frac{n\left(n-1\right)\left(n-2\right)!}{\left(n-2\right)!\times 2\times 1} \\ =\frac{n\left(n-1\right)}{2} \end{gathered}$$

Therefore, the number of handshakes possible is equal to $\frac{n\left(n-1\right)}{2}$.

c.

The number of different possible arrangements of n units is given by $n!$.

The total number of mathletes is 5.

The number of ways in which the five mathletes can be seated linearly is equal to:

$$\begin{gathered} 5!=5\times 4\times 3\times 2\times 1 \\ =120 \end{gathered}$$

In the case of a round table, if the mathletes shift once toward the right or left following the same order, the sitting arrangement does not change.

So, the number of ways computed above is divided by the total number of seats.

Thus, the number of ways in which the five mathletes can be seated, so that if there is a rotation the order of the seating arrangement does not change, is:

$$\begin{gathered} \frac{5!}{5}=\frac{5\times 4\times 3\times 2\times 1}{5} \\ =4! \\ =24 \end{gathered}$$

Therefore, the number of ways in which the five mathletes can be seated is equal to 24.

d.

If the number of mathletes is equal to n, the number of ways in which the n mathletes can be seated at a round table is equal to$\left(n-1\right)!$

Therefore, the number of ways in which the n mathletes can be seated is equal to$\left(n-1\right)!$