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virtus

  • 3 years ago

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  1. virtus
    • 3 years ago
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    \[nC4 + nC5\]

  2. akash123
    • 3 years ago
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    use d formula for n c r and do d simplification

  3. akash123
    • 3 years ago
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    nCr= n!/ r! (n-r)!

  4. virtus
    • 3 years ago
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    is it ok if you show me, kinda new to this

  5. akash123
    • 3 years ago
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    n!/ 4! (n-4)! + n!/ 5! (n-5)!

  6. akash123
    • 3 years ago
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    n(n-1)(n-2)(n-3)/24 + n(n-1)(n-2)(n-3)(n-4)/ 120

  7. akash123
    • 3 years ago
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    it's ok?

  8. virtus
    • 3 years ago
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    i still can't see it

  9. akash123
    • 3 years ago
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    see... n!= 1.2.3.4....n

  10. akash123
    • 3 years ago
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    right?

  11. virtus
    • 3 years ago
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    yes

  12. akash123
    • 3 years ago
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    n!/ (n-4)! = 1.2.3....(n-4)(n-3)(n-2)(n-1)n / (1.2.3....(n-4))

  13. virtus
    • 3 years ago
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    yeah...

  14. akash123
    • 3 years ago
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    so now u can solve ur problem....:)

  15. virtus
    • 3 years ago
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    let me try

  16. virtus
    • 3 years ago
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    n(n-1)(n-2)(n-3)/24 + n(n-1)(n-2)(n-3)(n-4)/ 120

  17. virtus
    • 3 years ago
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    i got that, but how do you simplify further?

  18. virtus
    • 3 years ago
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    LOL @hba looks like you have no choice but to help this damsel in distress

  19. hba
    • 3 years ago
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    @virtus i dont think so lol

  20. virtus
    • 3 years ago
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    16Cn +16C9 =17C9

  21. virtus
    • 3 years ago
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    FIND "n"

  22. virtus
    • 3 years ago
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    why not @hba

  23. hba
    • 3 years ago
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    Its been along time ive studied it and i dont want to get you into another confusion :(

  24. virtus
    • 3 years ago
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    OH LOL ok then

  25. akash123
    • 3 years ago
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    16Cn +16C9 =17C9 16Cn+16C9=16C9 + 16C8 16Cn=16C8 n=8

  26. akash123
    • 3 years ago
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    use the identity nCr + nC(r-1) = (n+1)Cr

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