To determine how many ways the 11 members of the club can line up for a photograph, we need to calculate the number of permutations of 11 distinct individuals.

The number of ways to arrange $n$ distinct items is given by $n!$ (n factorial).

Given $N = \{ \text{James, Sandy, Jane, Jenny, Bill, Tami, Kate, Joe, Allen, Ted, Carlos} \}$, we have $n = 11$.

So, the number of ways they can line up is:

$11! = 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$

Let's calculate this:

$11! = 39916800$

Therefore, the number of ways the 11 members can line up for a photograph is $39,916,800$ ways.

Do you want any details or have any questions? Here are some related questions that might interest you:

1. How many ways can 5 members out of the 11 be chosen and arranged in a line?
2. If one specific member must always be at the front, how many ways can the remaining 10 members line up?
3. How many ways can the 11 members be arranged if 2 specific members must be next to each other?
4. What if there are 11 positions but only 5 members need to be arranged in those positions, with the rest left empty?
5. How many ways can the 11 members be arranged if they must be in alphabetical order?

Tip: Remember that permutations are used when the order of selection matters, while combinations are used when the order does not matter.