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Convert the Following Products into Factorials: 1 · 3 · 5 · 7 · 9 ... (2n − 1)

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प्रश्न

Convert the following products into factorials:

1 · 3 · 5 · 7 · 9 ... (2n − 1)

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उत्तर

\[\ \left( 1 \right)\left( 3 \right)\left( 5 \right) . . . . . . . . . \left( 2n - 1 \right) = \frac{\left[ \left( 1 \right)\left( 3 \right)\left( 5 \right) . . . . . . . . . \left( 2n - 1 \right) \right]\left[ \left( 2 \right)\left( 4 \right)\left( 6 \right) . . . . . . . . . \left( 2n \right) \right]}{\left[ \left( 2 \right)\left( 4 \right)\left( 6 \right) . . . . . . . . . \left( 2n \right) \right]} \]
\[ = \frac{\left[ \left( 1 \right)\left( 2 \right)\left( 3 \right)\left( 4 \right)\left( 5 \right) . . . . . . . . . \left( 2n - 1 \right)\left( 2n \right) \right]}{2^n \left[ \left( 1 \right)\left( 2 \right)\left( 3 \right) . . . . . . . . . \left( n \right) \right]}\]
\[ = \frac{\left( 2n \right)!}{2^n n!}\]

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Factorial N (N!) Permutations and Combinations
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 16: Permutations - Exercise 16.1 [पृष्ठ ४]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 16 Permutations
Exercise 16.1 | Q 4.4 | पृष्ठ ४

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