In how many ways can 7 basketball players of different heights line up in a single row so that no player is standing between 2 people taller than she is?
To find the number of ways the basketball players can line up, we can use the concept of permutations. A permutation is an arrangement of objects in a specific order.
Let's approach this problem step by step:
Step 1: Identify the tallest player
Since no player can have two taller players standing on either side, we know that the tallest player must stand at one of the ends. Let's assume that the tallest player stands on the left end for now.
Step 2: Arrange the remaining players
The remaining 6 players are all shorter than the tallest player. We can think of them as just 6 other players. We can arrange them among themselves in any order, but we can't have a player taller than them between two players. So, there are 6! (6-factorial) possible arrangements of these players.
Step 3: Account for the tallest player's position
Since we had assumed the tallest player to be standing on the left end in step 1, we need to multiply the number of arrangements from step 2 by 2. This is because the tallest player can also stand on the right end.
Therefore, the total number of ways the 7 basketball players can line up is: 6! * 2 = 720 * 2 = 1440.
So, there are 1440 ways for the 7 basketball players to line up in a single row so that no player is standing between two people taller than she is.