SAT I Probability question

[Elbobo]

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?

 

[Dick]

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.

 

[Architeuthis Dux]

I can provide a logical and systematic approach to solving this problem. First, let's define the problem clearly. We have 5 cards laid out next to each other, and we want to find the number of different arrangements such that one of the inner cards is never on either end.

To solve this problem, we need to understand the concept of permutations. A permutation is an arrangement of objects in a specific order. In this case, our objects are the 5 cards and we want to arrange them in a specific order.

To find the number of arrangements, we can use the following formula: n! / (n-r)!, where n is the total number of objects and r is the number of objects we want to arrange. In this case, n=5 and r=3 (since we want to arrange 3 cards, leaving out the two end cards).

So, the number of arrangements can be calculated as 5! / (5-3)! = 5! / 2! = (5*4*3*2*1) / (2*1) = 120 / 2 = 60.

However, this includes arrangements where one of the inner cards is on either end. To exclude these arrangements, we need to subtract the number of arrangements where one of the inner cards is on either end. This can be calculated as 3! * 2 = 6 (since there are 3 possible positions for the inner card and 2 possible positions for the remaining card).

Therefore, the final number of arrangements is 60 - 6 = 54.

However, this still includes arrangements where the two inner cards are on either end. To exclude these arrangements, we need to subtract the number of arrangements where both inner cards are on either end. This can be calculated as 2! = 2 (since there are 2 possible positions for the two inner cards).

Therefore, the final number of arrangements is 54 - 2 = 52.

However, this still includes arrangements where the two inner cards are on either end and the remaining card is in the middle. To exclude these arrangements, we need to subtract the number of arrangements where the two inner cards are on either end and the remaining card is in the middle. This can be calculated as 1.

Therefore, the final number of arrangements is 52 - 1 = 51.

Hence, the total number of arrangements where one of the