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$\dashrightarrow \ \sf \dfrac{1}{n!} \ - \ \dfrac{1}{(n - 1)!} \ - \ \dfrac{1}{(n - 2)!}$
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Answer 1

Answer:

Question: To solve;

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$\dashrightarrow \ \sf \dfrac{1}{n!} \ - \ \dfrac{1}{(n - 1)!} \ - \ \dfrac{1}{(n - 2)!}$

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By using the factorial formula, We know that;

n! = n × (n - 1) × (n - 2)!

(n - 1)! = (n - 1) × (n - 2)!

Substitute these values in the expression given.

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$\sf \dashrightarrow \ \dfrac{1}{n(n - 1)(n - 2)!} \ - \ \dfrac{1}{(n - 1)(n - 2)!} \ - \ \dfrac{1}{(n - 2)!}$

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Since 1/[(n - 2)!] is common in all three fractions, let's take it out as a common factor.

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$\dashrightarrow \ \sf \dfrac{1}{(n - 2)!} \ \Bigg\{\dfrac{1}{n(n - 1)} \ - \ \dfrac{1}{(n - 1)} \ - \ 1 \Bigg\}$

Taking LCM;

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$\sf \dashrightarrow \ \dfrac{1}{(n - 2)!} \ \Bigg\{\dfrac{1}{n(n - 1)} \ - \ \Bigg(\dfrac{1}{(n - 1)} \times \dfrac{n}{n} \Bigg) \ - \ \Bigg( 1 \times \dfrac{n(n - 1)}{n(n - 1)} \Bigg) \Bigg\}$

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$\sf \dashrightarrow \ \dfrac{1}{(n - 2)!} \ \Bigg\{\dfrac{1}{n(n - 1)} \ - \ \dfrac{n}{n(n - 1)} \ - \ \dfrac{n(n - 1)}{n(n - 1)} \Bigg\}$

$\dashrightarrow \ \sf \dfrac{1}{(n - 2)!} \ \Bigg\{\dfrac{1 - n - n(n - 1)}{n(n - 1)} \Bigg\}$

$\dashrightarrow \ \sf \dfrac{1}{(n - 2)!} \ \Bigg\{\dfrac{1 - n - n^{2} + n}{n(n - 1)} \Bigg\}$

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$\dashrightarrow \ \sf \dfrac{1}{(n - 2)!} \ \Bigg\{\dfrac{1 - n^{2}}{n(n - 1)} \Bigg\}$

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On taking - out from (1 - n²) we get;

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$\dashrightarrow \ \sf \dfrac{1}{(n - 2)!} \ \Bigg\{\dfrac{-\big(-1 + n^{2}\big)}{n(n - 1)} \Bigg\}$

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$\dashrightarrow \ \sf \dfrac{1}{(n - 2)!} \ \Bigg\{\dfrac{-\big(n^{2} - 1\big)}{n(n - 1)} \Bigg\}$

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We know that;

(a - b)(a + b) = a² - b²

[Here, a = n and b = 1]

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$\boxed{\dashrightarrow \sf \dfrac{1}{(n - 2)!} \ \Bigg\{\dfrac{-\big(n + 1\big)\big(n - 1\big)}{n(n - 1)} \Bigg\}}$

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$\dashrightarrow \ \sf \dfrac{1}{(n - 2)!} \ \Bigg\{\dfrac{-\big(n + 1\big)}{n}\Bigg\}$

$\dashrightarrow \ \sf \dfrac{-\big(n + 1\big)}{n(n - 2)!}$

We won't be able to simplify it further, Hence solved.