# How many different ways can you get two heads and two tails in four tosses of a fair coin?

6

#### Explanation:

Let's list them out:

HHTT
HTTH
TTHH
THHT
HTHT
THTH

which is 6 ways.

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Another way to do this (and is actually how I caught that I'd missed 2 in the list above) is to work out the problem with a permutation formula (we care about the order of the coin flips, so that HHTT is different from TTHH. If they weren't different, like cards in a poker hand, then we'd be talking about combinations)

The general formula for a permutation is

P_(n,k)=(n!)/((n-k)!); n="population", k="picks"

Here we're saying that we have a population of 4 (taking each coin flip as a member of that population), picking 4 (we're doing a coin flip for all 4 of them). However, we need to adjust the formula a bit because we have only 2 results and each result appears twice. And so we divide by 2! for each of these groups (or in other words divide by 2!2!).

So we get:

(4!)/((4-4)!2!2!)=(4!)/((0!)(2!)(2!))=(4xx3xxcancel(2!))/((1)(cancel(2!))(2))=12/2=6