What is the total number of microstates possible to come out of 10 coinflips? This includes all microstates for each of the different macrostates.

The correct answer and explanation is:

The total number of microstates possible from 10 coin flips is: 210=10242^{10} = \boxed{1024}


Explanation (≈300 words):

In statistical mechanics and probability theory, a microstate is a specific detailed arrangement of a system’s components. For coin flips, a microstate is the exact sequence of heads (H) and tails (T) for each flip.

Each coin flip has two possible outcomes: heads (H) or tails (T). When flipping a coin 10 times, each flip is independent of the others. So, for each of the 10 flips, there are 2 possible outcomes. The total number of different combinations (microstates) is calculated by multiplying the outcomes: 2×2×⋯×2 (10 times)=210=10242 \times 2 \times \cdots \times 2 \text{ (10 times)} = 2^{10} = 1024

Each unique sequence, such as HTTHTHTHHH or TTTTTTTTTT, is a different microstate. This includes all sequences regardless of how many heads or tails they contain.

In contrast, a macrostate describes the system in terms of aggregate properties — in this case, how many heads (or tails) are in the sequence, regardless of order. For example, getting 5 heads and 5 tails is a macrostate. There are multiple microstates that correspond to this macrostate, specifically: (105)=252\binom{10}{5} = 252

This means 252 of the 1024 microstates have exactly 5 heads. Other macrostates (like 0 heads, 1 head, …, 10 heads) have different numbers of associated microstates (calculated using binomial coefficients).

If we sum the number of microstates for all possible macrostates (from 0 to 10 heads), we get: ∑k=010(10k)=210=1024\sum_{k=0}^{10} \binom{10}{k} = 2^{10} = 1024

Thus, the total number of microstates over all macrostates in 10 coin flips is: 1024\boxed{1024}

This concept illustrates a fundamental principle in probability and statistical mechanics — many microstates can correspond to a single macrostate.

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