A group of 9 workers make up a business team. One of the 9 is the manager, another member is the engineer. The remaining 7 are analysts. If the entire team is standing in line at the cafeteria on lunch, and the manager is standing 1st in line, and the engineer is standing 2nd, how many different orders can the line be in with the 7 analysts behind them?

Use factorials I think.

7! = 7 factorial = 7 * 6 * 5 * 4 * 3 * 2 * 1

That's what I was thinking, but do the 2 locked positions screw with my answer?

To find the number of different orders the line can be in with the given conditions, we can use the concept of permutations.

Permutations represent the different arrangements or orders in which a set of objects can be arranged. In this case, we want to find the number of permutations for a line of 7 analysts, excluding the manager and the engineer.

The number of permutations of a set of objects can be calculated using the formula:

nPr = n! / (n - r)!

Where n is the total number of objects, and r is the number of objects taken at a time.

In this case, we need to find the number of permutations of 7 analysts, so n = 7 and r = 7.

Plugging these values into the formula, we get:

7P7 = 7! / (7 - 7)!
= 7! / 0!
= 7! / 1
= 7!
= 7 x 6 x 5 x 4 x 3 x 2 x 1
= 5040

Hence, there are 5040 different orders the line can be in with the 7 analysts behind the manager and the engineer.