SAT I Probability question

In summary, for a sequence of 5 letters or cards, the total number of arrangements where one of the inner cards is never on either end is 72. This can be found by counting the possible positions for each card, starting with the inner card and moving outward, and multiplying them together.
  • #1
Elbobo
145
0

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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  • #2
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.
 
  • #3


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
 

Related to SAT I Probability question

What is the SAT I Probability question?

The SAT I Probability question is a type of question that appears on the SAT I exam, which is a standardized test used for college admissions in the United States. This question tests a student's understanding of probability and their ability to solve probability problems using mathematical concepts and formulas.

What concepts are typically covered in the SAT I Probability question?

The SAT I Probability question usually covers basic concepts such as independent and dependent events, permutations and combinations, and calculating probabilities using fractions and percentages. It may also include more advanced concepts such as conditional probability and expected value.

How can I prepare for the SAT I Probability question?

To prepare for the SAT I Probability question, it is important to review the basic concepts and formulas related to probability. Practice solving a variety of probability problems and make sure you understand the steps involved in solving them. Additionally, taking practice tests and reviewing the answers can help you become familiar with the types of questions that may appear on the SAT I exam.

Are there any tips for solving the SAT I Probability question?

One tip for solving the SAT I Probability question is to read the question carefully and identify the information given and what is being asked. Then, use the appropriate formula or method to solve the problem step-by-step. It is also helpful to make sure you understand the meaning of any key words or phrases in the question, such as "at least" or "or", as they can affect the way you approach the problem.

What is the importance of the SAT I Probability question?

The SAT I Probability question is important because it tests a student's ability to think critically and solve problems using mathematical concepts. It also assesses their ability to apply these concepts to real-world situations, which is a valuable skill in many fields of study and careers. A strong performance on the SAT I Probability question can also demonstrate a student's readiness for college-level coursework in mathematics.

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