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Home » \mathbfc = \mathbfa \times \mathbfb = \beginvmatrix \mathbfi & \mathbfj & \mathbfk \\ 2 & -3 & 1 \\ 1 & 4 & -2 \endvmatrix - AMAZONAWS

\mathbfc = \mathbfa \times \mathbfb = \beginvmatrix \mathbfi & \mathbfj & \mathbfk \\ 2 & -3 & 1 \\ 1 & 4 & -2 \endvmatrix - AMAZONAWS

📅 March 11, 2026 👤 scraface
Mar 11, 2026
\mathbf{c} = \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -3 & 1 \\ 1 & 4 & -2 \end{vmatrix}

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📌 Solution: First, calculate the total number of arrangements of the letters in "THEORETICAL". The word has 10 letters with repetitions: T (2), E (3), H (1), O (1), R (1), I (1), C (1), A (1). Total arrangements: $ \frac{10!}{2! \cdot 3!} = 302400 $. Next, subtract the arrangements where 'E' and 'A' are adjacent. Treat 'EA' or 'AE' as a single entity, reducing the problem to arranging 9 items (with T repeated twice and E reduced to 2). Number of such arrangements: $ 2 \cdot \frac{9!}{2! \cdot 2!} = 2 \cdot 362880 / 4 = 181440 $. The valid arrangements are $ 302400 - 181440 = 120960 $. Thus, the answer is $ \boxed{120960} $.Question: A 5 cm by 8 cm rectangle is inscribed in a circle. What is the number of centimeters in the circumference of the circle? Express your answer in terms of $\pi$.
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