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

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SUMMARY

The discussion focuses on calculating the number of arrangements of 5 distinct cards where one of the inner cards cannot occupy either end position. The correct answer is 72 arrangements. The solution involves determining the valid positions for the restricted card and calculating the permutations of the remaining cards accordingly. The method outlined uses basic combinatorial principles without the need for complex formulas.

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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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