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Question
Convert the following products into factorials:
1 · 3 · 5 · 7 · 9 ... (2n − 1)
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Solution
\[\ \left( 1 \right)\left( 3 \right)\left( 5 \right) . . . . . . . . . \left( 2n - 1 \right) = \frac{\left[ \left( 1 \right)\left( 3 \right)\left( 5 \right) . . . . . . . . . \left( 2n - 1 \right) \right]\left[ \left( 2 \right)\left( 4 \right)\left( 6 \right) . . . . . . . . . \left( 2n \right) \right]}{\left[ \left( 2 \right)\left( 4 \right)\left( 6 \right) . . . . . . . . . \left( 2n \right) \right]} \]
\[ = \frac{\left[ \left( 1 \right)\left( 2 \right)\left( 3 \right)\left( 4 \right)\left( 5 \right) . . . . . . . . . \left( 2n - 1 \right)\left( 2n \right) \right]}{2^n \left[ \left( 1 \right)\left( 2 \right)\left( 3 \right) . . . . . . . . . \left( n \right) \right]}\]
\[ = \frac{\left( 2n \right)!}{2^n n!}\]
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