~I Need Someone Who Understand Permutations And Combinations *Very well* ..~

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~I Need Someone Who Understand Permutations And Combinations *Very well* ..~

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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For what? I don't know if I qualify as *very well* but I know how they work. You got something really complicated?
@CliffSedge :They are not very Complicated but i want someone to help me through some questions ....
I'll at least take a look at one. I need to go in few minutes, though, but hopefully I can at least point you in the right direction.

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hmmm. As a start : IF : \[\Large C^{n}_{8} \times C^{n}_{6} \ge C^{n}_{7} \times C^{n}_{5}\] Prove That : \[\Large n \ge13\] _
Ok, so you know the combinations formula? \[nCk=\frac{n!}{k!(n-k)!}\]
yea
Ok, and as a start, n≥k so it has to be at least 8, just given the values of k.
It might also help to look at a simpler case using slightly smaller numbers to see how the factorials simplify. e.g. Show that 6C5 X 6C3 ≥ 6C4 X 6C2. That may give you some further insight.
Sorry, should have said "e.g. Show if ... ≥ ..." Because it might not.
:D
To be honest thats not giving me at least a small hint .. I was looking for a short way to end this question .. Anyway Ty for your help :)
If you play around with it a little bit, you'll see that the difference between n and k is important for the ultimate size of nCk - maybe also think of Pascal's Triangle . . .
Yeah, sorry, I'm feeling rushed, 'cause I need to go take care of some other matters, so the shortcut isn't coming to me at the moment. Good luck!
nvm ,i will find a way for it :) Ty again :)

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