Pascals Triangle: using it to solve 5(n_C_1) = (n + 1)_C_8
- Thread starter fencer817
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To learn how to simplify "<sub>m</sub>C<sub>n</sub>" (pronounced as "emm, choose enn"), try reviewing lessons online covering combinations.
. . . . .Google results for "combinations"
After you've learned the notation and have set up the exercise's equation, if you get stuck solving it, please reply showing how far you have gotten. Thank you.
Eliz.
. . . . .Google results for "combinations"
After you've learned the notation and have set up the exercise's equation, if you get stuck solving it, please reply showing how far you have gotten. Thank you.
Eliz.
Here, let's make it easy.
\(\displaystyle \L\\5C(n,1)-C((n+1),8)=0\)
\(\displaystyle \L\\\Rightarrow{\frac{-n^{8}}{40320}+\frac{n^{7}}{2016}-\frac{11n^{6}}{2880}+\frac{n^{5}}{72}-\frac{127n^{4}}{5760}+\frac{n^{3}}{288}+\frac{29n^{2}}{1120}+\frac{279n}{56}}=0\)
One of the roots of this is your answer.
\(\displaystyle \L\\5C(n,1)-C((n+1),8)=0\)
\(\displaystyle \L\\\Rightarrow{\frac{-n^{8}}{40320}+\frac{n^{7}}{2016}-\frac{11n^{6}}{2880}+\frac{n^{5}}{72}-\frac{127n^{4}}{5760}+\frac{n^{3}}{288}+\frac{29n^{2}}{1120}+\frac{279n}{56}}=0\)
One of the roots of this is your answer.