How many different card arrangements exclude inner cards at the ends?

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The discussion focuses on finding the number of arrangements for 5 different cards where one of the inner cards cannot be placed at either end. The solution involves recognizing that the restricted card can occupy one of three middle positions, allowing the remaining cards to be arranged around it. By calculating the available positions for each card sequentially, the total arrangements amount to 72. The method is illustrated using a similar example with letters, confirming the approach's validity. The final answer is established as 72 different arrangements.
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Homework Statement


How many different arrangements can be made with these 5 cards?
There are 5 different cards laid out next to each other. How many different arrangements can be made so that one of the inner cards is never on either end?




Homework Equations


None, probably


The Attempt at a Solution


I know the answer (72), but I have no idea how to get it. It's basic probability, so there should be a simple way of getting it (i.e., no formulas)...

I tried 252 minus 5 because I didn't know what else to do.

Can someone help guide me?
 
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Let's do the same problem with the sequence of letters ABCDE. You want all to count all of the arrangements where one of the letters, say B, doesn't occur at either end. Count the number of possibilities for each letter in turn. That means there are 3 different positions B can occupy (since it can't be at either end). That leaves 4 positions to put the A (since the B is in one of the places). Which in turn leaves 3 positions for the C (since A and B are occupying 2 positions). 2 positions for the D, and 1 for the E. Multiply them all together. 3*4*3*2*1=72.
 
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