A baseball team has 4 pitchers, who only pitch, and 16 other players, all of whom can play any position other than pitcher. For Saturday's game, the coach has not yet determined which 9 players to use nor what the batting order will be, except that the pitcher will bat last. How many different batting orders may occur?

2,075,673,600 batting orders may occur.

Step-by-step explanation:

The order of the first eight batters in the batting order is important. For example, if we exchange Jonathan Schoop with Adam Jones in the lineup, that is a different lineup. So we use the permutations formula to solve this problem.

Permutations formula:

The number of possible permutations of x elements from a set of n elements is given by the following formula:

First 8 batters

8 players from a set of 16. So

Last batter:

Any of the four pitchers.

How many different batting orders may occur?

4*518918400 = 2,075,673,600

2,075,673,600 batting orders may occur.

infinitely many solutions

step-by-step explanation:

we have been given the equation;

2(2x + 3) = -4 + 2(2x + 5)

we need to determine the value of x.

2(2x + 3), we expand this expression to obtain 4x + 6

similarly 2(2x + 5) will become 4x + 10

now we have;

4x + 6 = -4 + 4x + 10

4x + 6 = 4x + 10 - 4

4x + 6 = 4x + 6

the expression on the left hand side is identical to the one on the right hand side. therefore, we are looking at just a single straight line whose equation is;

y = 4x + 6

consequently, we shall have infinitely many solutions