\(\frac{100}{10!} = \frac{1}{8!} + \frac{x}{9!}\). Find x.(a) 1(b) 2(c) 3(d) 4This question was addressed to me by my school teacher while I was bunking the class.The origin of the question is Permutations-1 topic in section Permutations and Combinations of Mathematics – Class 11

\(\frac{100}{10!} = \frac{1}{8!} + \frac{x}{9!}\). Find x.(a) 1(b) 2(c

1.	\(\frac{100}{10!} = \frac{1}{8!} + \frac{x}{9!}\). Find x.(a) 1(b) 2(c) 3(d) 4This question was addressed to me by my school teacher while I was bunking the class.The origin of the question is Permutations-1 topic in section Permutations and Combinations of Mathematics – Class 11
Answer» Correct option is (a) 1 Easiest explanation: \(\frac{100}{10!} = \frac{1}{8!} + \frac{x}{9!}\). We know, n! = n.(n-1). (n-2). (n-3) …………… = n(n-1)! \(\frac{100}{10.9.8!} = \frac{1}{8!} + \frac{x}{9.8!}\) => \(\frac{100}{10.9} = \frac{1}{1} + \frac{x}{9}\) => \(\frac{10}{9} = 1 + \frac{x}{9}\) => \(\frac{x}{9} = \frac{10}{9} – 1 = \frac{1}{9}\) => x=1.

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\(\frac{100}{10!} = \frac{1}{8!} + \frac{x}{9!}\). Find x.(a) 1(b) 2(c) 3(d) 4This question was addressed to me by my school teacher while I was bunking the class.The origin of the question is Permutations-1 topic in section Permutations and Combinations of Mathematics – Class 11

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