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Eyad

  • 3 years ago

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

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  1. CliffSedge
    • 3 years ago
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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?

  2. Eyad
    • 3 years ago
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    @CliffSedge :They are not very Complicated but i want someone to help me through some questions ....

  3. CliffSedge
    • 3 years ago
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    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.

  4. Eyad
    • 3 years ago
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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 : \[\Large n \ge13\] _

  5. CliffSedge
    • 3 years ago
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    Ok, so you know the combinations formula? \[nCk=\frac{n!}{k!(n-k)!}\]

  6. Eyad
    • 3 years ago
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    yea

  7. CliffSedge
    • 3 years ago
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    Ok, and as a start, n≥k so it has to be at least 8, just given the values of k.

  8. CliffSedge
    • 3 years ago
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    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.

  9. CliffSedge
    • 3 years ago
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    Sorry, should have said "e.g. Show if ... ≥ ..." Because it might not.

  10. Eyad
    • 3 years ago
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    :D

  11. Eyad
    • 3 years ago
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    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 :)

  12. CliffSedge
    • 3 years ago
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    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 . . .

  13. CliffSedge
    • 3 years ago
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    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!

  14. Eyad
    • 3 years ago
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    nvm ,i will find a way for it :) Ty again :)

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