Question : $\frac{(100-1)(100-2)(100-3)....(100-200)}{100\times99\times98...3\times2\times1}$ is equal to:
Option 1: $\frac{100}{99\times98\times97\times...\times3\times2\times1}$
Option 2: $\frac{1}{99\times98\times97\times...\times3\times2\times1}$
Option 3: $0$
Option 4: $\frac{2}{99\times98\times97\times...\times3\times2\times1}$
Correct Answer: $0$
Solution :
Given: $\frac{(100-1)(100-2)(100-3)....(100-200)}{100\times99\times98...3\times2\times1}$
⇒ $\frac{(100-1)(100-2)(100-3).....(100-100).....(100-200)}{100\times99\times98...3\times2\times1}$
This will be equal to zero, as there will be a term in between $(100 – 100)$ equal to zero.
Hence, the correct answer is $0$.
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