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anonymous

  • one year ago

How do you apply Pascal's Triangle?

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  1. Nnesha
    • one year ago
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    To expand binomials you can apply pascal's triangle like you don't have to apply the binomial formula to solve \[\huge\rm ^4C_3\] just look at the triangle to find the coefficient of the terms 4th row http://www.mathwarehouse.com/animated-gifs/images/pascals-triangle-example-showing-recursion.gif

  2. Nnesha
    • one year ago
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    |dw:1440509714716:dw|

  3. Nnesha
    • one year ago
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    |dw:1440509928206:dw| 1+2 =3 (left side) 1+2 =3 right side you can also find these by using \[^nC_r =\frac{ n! }{ r!(n-r)! }\]formula

  4. Nnesha
    • one year ago
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    let me know if you hve any question cuz i think i didn't explain it clearly

  5. anonymous
    • one year ago
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    Wait... I didn't even know there was a formula...

  6. anonymous
    • one year ago
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    So how does the formula apply to the triangle?

  7. Nnesha
    • one year ago
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    tthe numbers in the pascal's triangle are the coefficient of binomial theorem \[\Large\rm(x+y)^4=\color{red}{1}x^4+\color{reD}{4}x^3y+\color{Red}{6}x^2y^2+\color{Red}{4}xy^3+\color{reD}{1}y^4\] here is an example red numbers are the coefficient you can find it by looking at the pascal's triangle or using that formula

  8. Nnesha
    • one year ago
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    let's say we nee 3rd term of 4th row \[^4C_3=\frac{ 4! }{ 3!(4-3)!}\] when you solve this you will get 6

  9. Nnesha
    • one year ago
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    do you know how to solve that^^?

  10. anonymous
    • one year ago
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    Ummm... yes?

  11. Nnesha
    • one year ago
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    no?

  12. Nnesha
    • one year ago
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    4!= 4 times 3 times 2 times 1 so 3!= ???

  13. Nnesha
    • one year ago
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    try it! :=)

  14. anonymous
    • one year ago
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    Hmmm. solve 3! ?

  15. Nnesha
    • one year ago
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    yeh 3! equal to what how would you expand 3! ?

  16. anonymous
    • one year ago
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    3*2*1

  17. Nnesha
    • one year ago
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    yes right \[^4C_3=\frac{\color{green}{ 4!} }{ \color{blue}{3!}\color{reD}{(4-3)!}}\]\[\frac{\color{green}{ 4 \times \cancel{3} \times \cancel{2} \times 1 }}{\color{blue}{\cancel{ 3} \times \cancel{2} \times 1}\color{red}{(1)}!}\] answer would be one

  18. Nnesha
    • one year ago
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    yes right \[^4C_3=\frac{\color{green}{ 4!} }{ \color{blue}{3!}\color{reD}{(4-3)!}}\]\[\frac{\color{green}{ 4 \times \cancel{3} \times \cancel{2} \times 1 }}{\color{blue}{\cancel{ 3} \times \cancel{2} \times 1}\color{red}{(1)}!}\] answer would be 4

  19. Nnesha
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @Nnesha let's say we nee 3rd term of 4th row \[^4C_3=\frac{ 4! }{ 3!(4-3)!}\] when you solve this you will get 6 \(\color{blue}{\text{End of Quote}}\) my bad i meant you will get 4

  20. anonymous
    • one year ago
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    You get 4? Now I am confused...

  21. Nnesha
    • one year ago
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    what did you get ?? :(

  22. Nnesha
    • one year ago
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    |dw:1440513870886:dw|

  23. anonymous
    • one year ago
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    Ohh... 4th term... I got 6... I looked at the 3rd term.

  24. Nnesha
    • one year ago
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    yeah i didn't know that either that's why i said 6 but it starts from 0

  25. Nnesha
    • one year ago
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    |dw:1440514055424:dw|

  26. anonymous
    • one year ago
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    Oh... oops.

  27. Nnesha
    • one year ago
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    so you can find coefficients of binomial by using that formula or remember the pattern :=)

  28. anonymous
    • one year ago
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    Ok, Thanks. I think I get it.

  29. Nnesha
    • one year ago
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    np :=)

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