Katrina, Sarah, Dawn, Sergio, Eduardo, Maria, and lan have all been invited to a dinner party. They arrive randomly and each person arrives at a different time.
a. In how many ways can they arrive?
b. In how many ways can Katrina arrive first and lan last?
c. Find the probability that Katrina will arrive first and lan last.
a. $\square$ (Type an integer.)
b. $\square$ (Type an integer.)
c. $\square$ (Type a fraction. Simplify your answer.)
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Step 1 ：First, we need to find the total number of ways the seven people can arrive. This is a permutation problem, as the order in which the people arrive matters. The total number of ways is given by the factorial of the number of people, which is \(7!\).

Step 2 ：Using the factorial function, we find that \(7! = 5040\). So, there are 5040 different ways the seven people can arrive.

Step 3 ：Next, we need to find the number of ways Katrina can arrive first and Ian can arrive last. This is also a permutation problem, but now we only have 5 people whose order of arrival can vary, as the positions of Katrina and Ian are fixed. The number of ways is given by the factorial of the number of people, which is \(5!\).

Step 4 ：Using the factorial function, we find that \(5! = 120\). So, there are 120 different ways Katrina can arrive first and Ian can arrive last.

Step 5 ：Finally, we need to find the probability that Katrina will arrive first and Ian will arrive last. This is given by the ratio of the number of ways Katrina can arrive first and Ian can arrive last to the total number of ways the seven people can arrive. So, the probability is \(\frac{120}{5040}\).

Step 6 ：Simplifying the fraction, we find that the probability is \(\frac{1}{42}\).

Step 7 ：So, the final answers are: a. \(\boxed{5040}\), b. \(\boxed{120}\), c. \(\boxed{\frac{1}{42}}\).