How Many Possible Arrangements of 10 Tossed Coins Include 5 Heads and 5 Tails?

The problem of determining the number of possible arrangements of 10 tossed coins, where exactly 5 are heads (H) and 5 are tails (T), is a classic example in combinatorial mathematics. This type of problem often arises in probability theory and statistical analysis. Here, we explore various methods to solve this problem, including sophisticated programming techniques and mathematical principles.

Introduction

When we toss a fair coin 10 times, each toss has two possible outcomes: heads or tails. Therefore, the total number of possible outcomes when tossing the coin 10 times is simply (2^{10}), which is 1024. However, the question specifically asks for the number of outcomes with exactly 5 heads and 5 tails. This can be determined using the binomial coefficient formula and combinatorial principles.

Mathematical Solution

The number of ways to choose 5 heads out of 10 coin tosses can be calculated using the binomial coefficient formula. The formula is given by:

[{n choose k} frac{n!}{k!(n-k)!}]

In this problem, (n 10) (the total number of coin tosses) and (k 5) (the number of heads). Plugging these values into the formula, we get:

[{10 choose 5} frac{10!}{5!5!} 252]

This means there are 252 distinct ways to get exactly 5 heads and 5 tails in 10 coin tosses.

Programming Solution

A program can be written to compute the number of possible arrangements. One such implementation could use the J programming language or a statistical programming language like R. Here, we provide an example using R:

```r # Using the R programming language library(combinat) my_list This R script generates all possible permutations of 5 heads and 5 tails in 10 coin tosses. Running the script will confirm that there are indeed 252 unique permutations.

Alternative Approach

Another way to approach this problem is through the use of the binomial probability formula. The probability of obtaining exactly 5 heads in 10 tosses of a fair coin is given by:

[P(X 5) {10 choose 5} times 0.5^5 times 0.5^5 {10 choose 5} times 0.5^{10}]

Simplifying, we get:

[P(X 5) 252 times 0.5^{10} 252 times frac{1}{1024} frac{252}{1024} approx 0.2461]

This probability can then be used to calculate the number of possible arrangements by multiplying the probability by the total number of outcomes (1024). The result is 252, confirming our earlier calculation.

Conclusion

In conclusion, the number of possible arrangements of 10 tossed coins resulting in exactly 5 heads and 5 tails is 252. This can be determined mathematically using the binomial coefficient formula, verified through programming, or calculated using the binomial probability approach. Regardless of the method used, the final answer remains the same, reflecting the fundamental principles of combinatorial mathematics.