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Simplify 1n!-1(n-1)!-1(n-2)!

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प्रश्न

Simplify `1/("n"!) - 1/(("n" - 1)!) - 1/(("n" - 2)!)`

बेरीज
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उत्तर

`1/("n"!) - 1/(("n" - 1)!) - 1/(("n" - 2)!)`

= `1/("n"("n" - 1)("n" - 2)!) - 1/(("n" - 1)("n" - 2)!) - 1/(("n" - 2)!)`

= `1/(("n" - 2)!) [1/("n"("n" - 1)) - 1/(("n" - 1)) - 1]`

= `1/(("n" - 2)!) [(1 - "n" - "n"("n" - 1))/("n"("n" - 1))]`

= `(1 - "n"^2)/("n"("n" - 1)("n" - 2)!)`

= `(-("n" - 1)("n" + 1))/("n"("n" - 1)("n" - 2)!)`

= `(-("n" + 1))/("n"("n" - 2)!)`

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पाठ 3: Permutations and Combination - Exercise 3.2 [पृष्ठ ५०]

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बालभारती Mathematics and Statistics (Arts and Science) Part 2 [English] Standard 11 Maharashtra State Board
पाठ 3 Permutations and Combination
Exercise 3.2 | Q 10. (iii) | पृष्ठ ५०

संबंधित प्रश्‍न

Evaluate: 10!


Evaluate: 10! – 6!


Compute: `(12!)/(6!)`


Compute: `(12/6)!`


Compute: (3 × 2)!


Compute: 3! × 2!


Compute: `(8!)/(6! - 4!)`


Compute: `(8!)/((6 - 4)!)`


Write in terms of factorial.

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Write in terms of factorial.

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Write in terms of factorial.

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Evaluate : `("n"!)/("r"!("n" - "r")!)` for n = 15, r = 8


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