A coin is flipped ten times, resulting in either heads or tails on each flip. How many possible outcomes contain at most three tails?

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James Kilian · Accepted Answer

Step 1&amp;nbsp;Definitions&amp;nbsp;If an event can happen in m ways and a subsequent event can happen in n ways, the product rule applies, then the number of ways that the two events can occur in sequence is then m*n.&amp;nbsp;Definition permutation (order is important):&amp;nbsp;P(n,r)=n!(n−r)!&amp;nbsp;Definition combination (order is not important):&amp;nbsp;C(n,r)=(nr)=n!r!(n−r)!&amp;nbsp;with n!=n*(n-1)*...*2*1.&amp;nbsp;Step 2&amp;nbsp;The order of the heads/tails is not important (since we are interested in the number of heads, not the order of the heads), thus we need to use the definition of combination.&amp;nbsp;n=10&amp;nbsp;r≤2&amp;nbsp;Consider the following when defining a combination:C(10,0)=10!0!(10−0)!=10!0!10!=1&amp;nbsp;C(10,1)=10!1!(10−1)!=10!1!9!=10&amp;nbsp;C(10,2)=10!2!(10−2)!=10!2!8!=45&amp;nbsp;C(10,3)=10!3!(10−3)!=10!3!7!=120&amp;nbsp;Add the number of outcomes for each value of r:&amp;nbsp;120+45+10+1=176

A coin is flipped 10 times where each flip comes

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