THE TICKET COUNTER LINE

Suppose 50 people are standing in a single line in front of a ticket counter. How many ways are there for the 50 people to show up?

The first thing I wrote down for this problem was U₅₀ = 50⁵⁰. This reveals that I was thinking that there were 50 repeatable choices for each position in line. This was not a realistic point-of-view, but I dismissed this problem very quickly and didn't realize my mistake until it was discussed in class. Then it was very obvious to me that U₅₀ = 50! Because: If you have 50 choices for the first spot in line, once that spot is taken, there are only 49 people left from which to choose the person in the second spot, etc... Once you've realized this decreasing number of options (one less each time), and once you realize that those numbers need to be multiplied (For each choice for #1, there are 49 choices for #2, so there are 50 * 49 choices for one and two, etc.) The factorial notation quickly gives you a closed formula: U_n = n! I can't think of a solution on a more intuitive level.

Even the recursive formula isn't particularly challenging. If U_n-1 is the product of all the terms from 1 to n-1. Then U_n is merely U_n-1 times the new n. Or: U_n = U_n-1 * n.

I don't know what else to say about this problem. It is different from most of the others in this set because it deals with the product of a sequence of numbrs as opposed to the sum of a sequence of numbers.

SUMMARY OF SOLUTIONS:

Numerical Solution

U₅₀ = 50! = 3.0414e+64

Recursive formula:

U_n = U_n-1 *n

Closed formula:

U_n = n!