Try This Combinatorics Problem

There are 8 possible outcomes.

We can list the outcomes out as HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.

Alternatively, we can use the rule of product. Since there are 2 possibilities for each coin, there are a total of $2 \times 2 \times 2 = 8$ possible outcomes.

We can think of this recursively to derive a generalization of the problem:

If we flip one coin, there are two outcomes: Heads or Tails.

If we flip two coins, we can think of this as flipping the first coin (so two outcomes) and then flipping another coin where there are another two outcomes. Laws of probability tell us that we should multiply two by two to get four outcomes if we flip two coins.

Flipping three coins is the same as flipping two coins and then flipping another coin, so we get $2 \times 4 = 8$ outcomes.

To generalize, let $a_n$ be the number of possible outcomes with $n$ coin flips. Then this follows the recursive sequence $a_n = 2 a_{n-1}$ , with $a_1 = 2$ . Using this, we find the explicitly defined sequence $a_n = 2^n$ .

There are 8 possibilities when 3 coins are flipped I.e., HHH, HHT, HTT, TTH, THH, HTH, THT, TTT

(H + T)^ 3 = H H H + 3 H H T + 3 H T T + T T T for 8 outcomes;

(0.5 H + 0.5 T)^ 3 = (H^3 + 3 H^2 T + 3 H T^2 + T^3)/ 8 for a sum of 1 where 0.5 could become 0.49 or 0.51.