The coach of a certain Philadelphia basketball team must arrange 5 centers and 2 point guards in a row of 7 seats inside a tank. If the first and last seats in the tank must be occupied by a point guard, how many different ways can the players be arranged in the row?
This problem asks us about an arrangement of things where order matters, which tells us that this is a permutation problem. Permutation problems can be solved using a method we call the “slot method.” This method has us draw a blank for each choice we need to make. In this case, we have 7 seats to fill in our unstoppable tank, so we will draw 7 blanks:
___ ___ ___ ___ ___ ___ ___
Each blank will represent one our seats. Now, we want to fill in the number of options we have for each seat. Since the first seat must have one of our two point guards, we put a 2 in our first blank. The last seat must also have a point guard, but since we have already assigned one of our point guards to the first seat, we only have one choice for the last seat. We then put a 1 in the last blank.
_2_ ___ ___ ___ ___ ___ _1_
Now for each of the remaining seats, we must have a center. Since we have 5 centers, we have 5 different choices for who sits in the second seat. We put a 5 in the second blank. Once a center is assigned to the second seat, we have 4 centers remaining to choose from for the third seat. We put a 4 in the third blank. This trend continues until we our blanks appear as follows:
_2_ _5_ _4_ _3_ _2_ _1_ _1_
Once all blanks are filled in, we simply multiply all of our numbers together:
Thus, there are 240 different ways to seat our seven players in the tank.
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