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# Problem I could not solved.

(n-3)(n-4)!/(n-5)(n-6)!=?

Note by Mehdi Balti
1 year, 3 months ago

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Comment deleted Jul 20, 2015

Can you explain it :) · 1 year, 3 months ago

@Mehdi Balti, could you please clarify what the problem is? Thanks. · 1 year, 3 months ago

It's factorial and the answer will be in positive integer.Not in decimal (n-3)(n-4)!/(n-5)(n-6)!=? · 1 year, 3 months ago

Based on how you've clarified the problem, I'm assuming you wish to find the value of

$x = \frac{(n-4)!(n-3)}{(n-6)!(n-5)},$

where $$x$$ is a positive integer.

Note that

\begin{align} x &= \frac{(n-4)!(n-3)}{(n-6)!(n-5)} \nonumber \\ &= \frac{(n-3)[(n-4)(n-5)(n-6)\cdots(3)(2)(1)]}{(n-5)[(n-6)(n-7)(n-8)\cdots(3)(2)(1)]} \nonumber \\ &= \frac{(n-3)(n-4)(n-5)!}{(n-5)!} \nonumber \\ &= (n-3)(n-4) \nonumber \end{align}

This value is different for different values of $$n$$. I'm not sure if this is what you want, although $$(n-3)(n-4)$$ is clearly a positive integer for positive integer values of $$n$$. Please clarify further. Thanks! · 1 year, 3 months ago

Yes n value is positive integer.Really thanks @Victor Loh I got it completely Now :) Thumps up for u Cheers :) · 1 year, 3 months ago