Puzzles

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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

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.

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