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Ten identical coins are tossed. How many possible arrangements of the coins include five heads and five tails?

Sagot :

Sure, let's break down the problem step by step:

1. Understand the Problem:
- We have 10 identical coins.
- We need to determine how many ways we can arrange these coins so that we get exactly 5 heads and 5 tails.

2. Identify the Concept:
- This is a classic problem of combinations where we want to choose 5 positions out of 10 for either heads or tails. The remaining positions will automatically be filled by the other coin face.

3. Apply the Combinatorial Formula:
- The formula for combinations is given by:
[tex]\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \][/tex]
- Here, [tex]\(n\)[/tex] is the total number of coins, which is 10, and [tex]\(k\)[/tex] is the number of positions we are choosing for heads (or tails), which is 5.

4. Calculate the Number of Combinations:
- Using the combination formula:
[tex]\[ \binom{10}{5} = \frac{10!}{5! \cdot 5!} \][/tex]
This formula gives us the total number of ways to choose 5 positions out of 10 for heads.

5. Result:
- After performing the calculation (which involves factorials), we find that:
[tex]\[ \binom{10}{5} = 252 \][/tex]

So, the number of possible arrangements of the 10 coins that result in exactly 5 heads and 5 tails is 252.