Question

1. (12 points) The tennis team at Taft High School has 8 players, and the tennis team at McKinley High School has 7 players. Rebecca is on the team at Taft, and her sister Leah is on the team at McKinley. The two teams are going to play a tournament with 4 rounds. In each round, one player from Taft will play a match against one player from McKinley. Each player can play in at most one match. A schedule for the tournament consists of an ordered list of the names of the players from each school who will play in each of the four rounds. The schedule does not include the results of each match. (a) How many schedules are there? (b) How many schedules include neither sister? (c) How many schedules include both sisters and have them playing against each other? (d) How many schedules include both sisters, but have them playing in different matches?

Answer 1

Answer:

(a) 1,411,200.

(b) 302,400.

(c) 100,800.

(d) 1,008,000.

Step-by-step explanation:

Permutation is the number of ways to select k items from n distinct items in a specific order.

The formula to compute the permutation or arrangement of k items is:

$^{n}P_{k}=\frac{n!}{(n-k)!}$

(a)

Compute the total number of schedules as follows:

$\text{Total Number of Schedules}=\ ^{8}P_{4}\times \ ^{7}P_{4}$

                                          $=\frac{8!}{(8-4)!}\times \frac{7!}{(7-4)!}\\\\=\frac{8\times 7\times 6\times 5\times 4!}{4!}\times \frac{ 7\times 6\times 5\times 4\times 3!}{3!}\\\\=(8\times 7\times 6\times 5)\times (7\times 6\times 5\times 4)\\\\=1411200$

Thus, the total number of schedules are, 1,411,200.

(b)

Compute the number of schedules that includes neither sister as follows:

If neither of sister are included in the schedule then the number of players from the team are 7 and 6 respectively.

$\text{Number of Schedules without the sisters}=\ ^{7}P_{4}\times \ ^{6}P_{4}$

                                                               $=\frac{7!}{(7-4)!}\times \frac{6!}{(6-4)!}\\\\=\frac{ 7\times 6\times 5\times 4\times 3!}{3!}\times \frac{ 6\times 5\times 4\times 3\times 2!}{2!}\\\\=(7\times 6\times 5\times 4)\times (6\times 5\times 4\times 3)\\\\=302400$

Thus, the number of schedules that includes neither sister is 302,400.

(c)

Compute the number of schedules including both sisters and have them playing against each other as follows:

If both the sisters are playing against each other then there will be three more places to be filled in both the teams.

And the two sisters can play in any of the 4 matches.

$\text{number of schedules including} \\\text{both sisters against each other}=(\ ^{7}P_{3}\times \ ^{6}P_{3})\times 4$

                                                $=[\frac{7!}{(7-3)!}\times \frac{6!}{(6-3)!}]\times 4\\\\=[\frac{ 7\times 6\times 5\times 4!}{4!}\times \frac{ 6\times 5\times 4\times 3!}{3!}]\times 4\\\\=[(7\times 6\times 5)\times (6\times 5\times 4)]\times 4\\\\=100800$

Thus, the number of schedules including both sisters and have them playing against each other is 100,800.

(d)

Compute the number of schedules including both sisters as follows:

$\# \text {Schedules with the Sisters}=\text{Total}\ \#\ \text{Schedules }-\ \# \text {Schedules without the Sisters}$                                             $=1411200-302400\\=1108800$

Now compute the number of schedules having the sisters playing in different matches as follows:

$\text{Playing in Different Matches}=\# \text {Schedules with the Sisters}-\text{Playing together}$

                                              $=1108800-100800\\=1008000$

Thus, the number of schedules including both sisters, but have them playing in different matches is 1,008,000.