Could someone please explain how one gets to the second part of this equation?
Thanks.
$$\left({\frac{\left(\begin{pmatrix}
{\mathtt{7}}
\\
{\mathtt{2}}
\end{pmatrix}
{\mathtt{\,\times\,}}\begin{pmatrix}
{\mathtt{5}}
\\
{\mathtt{4}}
\end{pmatrix}
\right)}{\begin{pmatrix}
{\mathtt{12}}
\\
{\mathtt{6}}
\end{pmatrix}
}}\right) = {\frac{\left({\mathtt{21}}{\mathtt{\,\times\,}}{\mathtt{5}}\right)}{{\mathtt{924}}}} = \tiny\text{Error: }$$
+23254 The first section can be written as 7C2 · 5C4 / 12C6
7C2 is evaluated as 7! / (2! · 5!) which can be reduced to 7·6 / (2·1) = 42 / 2 = 21.
5C4 is evaluated as 5! / (4! · 1!) which can be reduced to 5 / (1) = 5.
12C6 is evaluated as 12! / (6! · 6!) which can be reduced to 11·3·4·7 / (1) = 924.
In general, nCr becomes n! / ( r! · (n - r)! ), so write 12! / (6! · 6!) as
(12·11·10·9·8·7·6·5·4·3·2·1) / ( 6·5·4·3·2·1 · 6·5·4·3·2·1 ) and then cancel like mad!
Questions?
+23254 The first section can be written as 7C2 · 5C4 / 12C6
7C2 is evaluated as 7! / (2! · 5!) which can be reduced to 7·6 / (2·1) = 42 / 2 = 21.
5C4 is evaluated as 5! / (4! · 1!) which can be reduced to 5 / (1) = 5.
12C6 is evaluated as 12! / (6! · 6!) which can be reduced to 11·3·4·7 / (1) = 924.
In general, nCr becomes n! / ( r! · (n - r)! ), so write 12! / (6! · 6!) as
(12·11·10·9·8·7·6·5·4·3·2·1) / ( 6·5·4·3·2·1 · 6·5·4·3·2·1 ) and then cancel like mad!
Questions?
