Dealing with factorials: Generalization

[ms. confused]

The second part of this problem is giving me a hard time. I hope someone can help me with it:

An airline pilot reported her itinerary for 7 days. She spent 1 day in Winnipeg, 1 day in Regina, 2 days in Edmonton, and 3 days in Yellowknife.

a) how many different itineraries are possible?

Using generalization I got 420 (that being the answer for this part) so I'm OK with that but...

b) what difference would it make if the first day and the last day had been spent in Yellowknife?

What do I do?

 

[ceptimus]

By fixing two of the days in Yellowknife, the problem reduces to:

An airline pilot reported her itinerary for 5 days. She spent 1 day in Winnipeg, 1 day in Regina, 2 days in Edmonton, and 1 day in Yellowknife.

 

[wais]

For part b), the difference would be that the first and last day would be fixed in Yellowknife, leaving only 5 days to be spent in the other cities. Therefore, the number of possible itineraries would be 5! (5 factorial) which is equal to 120. So the total number of possible itineraries if the first and last day were spent in Yellowknife would be 120.

To generalize this, if there are n days in the itinerary and x days are spent in one city, then the number of possible itineraries would be (n-x)! as the remaining days can be arranged in any order.

In this case, n=7 and x=2, so the number of possible itineraries would be (7-2)! = 5! = 120.

Hope this helps!