anonymous
  • anonymous
How do you apply Pascal's Triangle?
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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Nnesha
  • Nnesha
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
Nnesha
  • Nnesha
|dw:1440509714716:dw|
Nnesha
  • Nnesha
|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

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Nnesha
  • Nnesha
let me know if you hve any question cuz i think i didn't explain it clearly
anonymous
  • anonymous
Wait... I didn't even know there was a formula...
anonymous
  • anonymous
So how does the formula apply to the triangle?
Nnesha
  • Nnesha
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
Nnesha
  • 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
Nnesha
  • Nnesha
do you know how to solve that^^?
anonymous
  • anonymous
Ummm... yes?
Nnesha
  • Nnesha
no?
Nnesha
  • Nnesha
4!= 4 times 3 times 2 times 1 so 3!= ???
Nnesha
  • Nnesha
try it! :=)
anonymous
  • anonymous
Hmmm. solve 3! ?
Nnesha
  • Nnesha
yeh 3! equal to what how would you expand 3! ?
anonymous
  • anonymous
3*2*1
Nnesha
  • Nnesha
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
Nnesha
  • Nnesha
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
Nnesha
  • Nnesha
\(\color{blue}{\text{Originally Posted by}}\) @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
anonymous
  • anonymous
You get 4? Now I am confused...
Nnesha
  • Nnesha
what did you get ?? :(
Nnesha
  • Nnesha
|dw:1440513870886:dw|
anonymous
  • anonymous
Ohh... 4th term... I got 6... I looked at the 3rd term.
Nnesha
  • Nnesha
yeah i didn't know that either that's why i said 6 but it starts from 0
Nnesha
  • Nnesha
|dw:1440514055424:dw|
anonymous
  • anonymous
Oh... oops.
Nnesha
  • Nnesha
so you can find coefficients of binomial by using that formula or remember the pattern :=)
anonymous
  • anonymous
Ok, Thanks. I think I get it.
Nnesha
  • Nnesha
np :=)

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