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I'm revising for an exam on permutation. I saw this question

Find a positive integer n such that P (n, 2) = 110.

The answer

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I don't understand how n! became (n-1)n when we don't know what n is. I need help. Thanks

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    Do you understand that $n! = (n-2)!(n-1)n$ (assuming of course $n \ge 2$)? Then cancel the $(n-2)!$ in numerator and denominator. – 2017-01-24
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    yes. but why are we assuming its >= 2. I'm not getting it yet @RobertIsrael – 2017-01-24
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    @Calypso: clearly $n=0,1$ don't work, so $n$ must be at least $2$ – 2017-01-24
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    Ooooh, it feels like a sudden state of epiphany. I understand now. Thanks @RobertIsrael – 2017-01-24

3 Answers 3

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$$n!=(n)(n-1)!=(n)(n-1)(n-2)!$$

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For integers, factorials have the property

$$ n! = n \cdot (n - 1)! $$

Hence you have $n = n\cdot (n-1)! = n(n-1)\cdot (n-2)!$.

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Simply take example - You have 5!

Then 5! = 5 × 4!

We stop expanding it up to 4!

We subtract 1 from 5 to get 4!

Suppose you have n instead of 5 then you subtract 1 from n. So we have now (n-1)!

Therefore,

n! = n × (n - 1)!

If further want to expand subtract 1 from factorial term again.

n! = n × (n - 1) × (n - 2)!

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    Still doubt please ask. – 2017-01-24