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Prove that the Product of 2n Consecutive Negative Integers is Divisible by (2n)!

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प्रश्न

Prove that the product of 2n consecutive negative integers is divisible by (2n)!

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उत्तर

Let  

\[2n\] negative integers be 
\[\left( - r \right), \left( - r - 1 \right), \left( - r - 2 \right), . . . . , . . . , \left( - r - 2n + 1 \right)\]
Then, product = \[\left( - 1 \right)^{2n} \left( r \right)\left( r + 1 \right)\left( r + 2 \right), . . . . , . . . \left( r + 2n - 1 \right)\]
\[= \frac{\left( r - 1 \right)! \left( r \right)\left( r + 1 \right) \left( r + 2 \right) . . . . . . \left( r + 2n - 1 \right)}{\left( r - 1 \right)!}\]
\[ = \frac{\left( r + 2n - 1 \right)!}{\left( r - 1 \right)!}\]
\[ = \frac{\left( r + 2n - 1 \right)!}{\left( r - 1 \right)!\left( 2n \right)!} \times \left( 2n \right)!\]
\[ = {}^{r + 2n - 1} C_{2n} \times \left( 2n \right)!\]
This is divisible by 
\[\left( 2n \right)! .\]
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Factorial N (N!) Permutations and Combinations
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अध्याय 17: Combinations - Exercise 17.1 [पृष्ठ ८]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 17 Combinations
Exercise 17.1 | Q 16 | पृष्ठ ८

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