Advent calendar 2017
15 December
The string ABBAABBBBB is 10 characters long, contains only A and B, and contains at least three As.
Today's number is the number of different 10 character strings of As and Bs that have at least three As.
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The number of 10 character strings containing at least 3 As is equal to
the number of 10 character strings containing exactly 3 As
plus
the number of 10 character strings containing exactly 4 As
plus ... plus
the number of 10 character strings containing exactly 10 As.
The number of 10 character strings containing exactly \(a\) As is \(\left(\begin{array}{c}10\\a\end{array}\right)\) ("10 choose \(a\)").
Therefore the number we are looking for is
$$
\left(\begin{array}{c}10\\3\end{array}\right)
+
\left(\begin{array}{c}10\\4\end{array}\right)
+
\left(\begin{array}{c}10\\5\end{array}\right)
+
\left(\begin{array}{c}10\\6\end{array}\right)
+
\left(\begin{array}{c}10\\7\end{array}\right)
+
\left(\begin{array}{c}10\\8\end{array}\right)
+
\left(\begin{array}{c}10\\9\end{array}\right)
+
\left(\begin{array}{c}10\\10\end{array}\right)
$$
You can work this out by either using the formula \(\left(\begin{array}{c}n\\r\end{array}\right)=\frac{n!}{r!(n-r)!}\), or by remembering that
these numbers all appear in the 10th row of
Pascal's triangle.
The answer is 968.
