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Given the following text description, write Python code to implement the functionality described below step by step
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Description:
 Ejercicio
Simplifica los cocientes entre factoriales
Step1:  Ejercicio
Calcula las siguientes operaciones
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Python Code:
enunciado = list([r'\frac{7!}{6!}',r'\frac{{8!}}{{9!}}',r'\frac{{9!}}{{5!\cdot 4!}}',r'\frac{{m!}}{{(m - 1)!}}', r'\frac{{( {m + 1} )!}}{{( {m - 1} )!}}'])
enunciado
enunciado = list([r'\frac{7!}{6!}',r'\frac{{8!}}{{9!}}',r'\frac{{9!}}{{5!\cdot 4!}}',r'\frac{{m!}}{{(m - 1)!}}', r'\frac{{( {m + 1} )!}}{{( {m - 1} )!}}'])
enunciado_sympy=[]
for i in enunciado : 
    enunciado_sympy.append(parse_latex(i)); 
    
enunciado_sympy
for i in range(len(enunciado_sympy)) : 
    display(md("$"+enunciado[i]+"  \\rightarrow   "+latex(simplify(enunciado_sympy[i]))+"$"))
    
Explanation: Ejercicio
Simplifica los cocientes entre factoriales:
 - $\frac{7!}{6!}$
 - $\frac{{8!}}{{9!}}$
 - $\frac{{9!}}{{5!.4!}}$     
 - $\frac{{m!}}{{(m - 1)!}}$      
 - $\frac{{\left( {m + 1} \right)!}}{{\left( {m - 1} \right)!}}$
End of explanation
from sympy.functions.combinatorial.numbers import nC, nP, nT
nC(5,3)
from sympy import *
expr = sympify("nC(5,3)")
display(expr.expand())
enunciado = [[252,250], [25,3], [25,4]]
for i in range(len(enunciado)):
    display(nC(enunciado[i][0],enunciado[i][1]))
nC(enunciado[0][0],enunciado[0][1])
factorial(252)/(factorial(250)*factorial(2))
Explanation: Ejercicio
Calcula las siguientes operaciones:
- $\binom{252}{250}$  
- $\binom{25}{3} + \binom{25}{4} = \binom{26}{4}$
- $\binom{9}{6} + \binom{9}{7} + \binom{10}{2}=\binom{10}{7}+\binom{10}{8}=\binom{11}{8}$
- $\binom{4}{2} + \binom{4}{3} + \binom{5}{4}+\binom{6}{5} + \binom{7}{6} + \binom{8}{7}=\binom{9}{7}$
- $\binom{4}{0} + \binom{4}{1} + \binom{4}{2}+\binom{4}{3} = 2^4-1$
\endmatrix } \right)\, + \,\left( {\matrix
   {25}  \ 
   4  \ 
\endmatrix } \right)$$ iii) $$\left( {\matrix
   9  \ 
   6  \ 
\endmatrix } \right)\, + \,\left( {\matrix
   9  \ 
   7  \ 
\endmatrix } \right)\, + \,\left( {\matrix
   {10}  \ 
   2  \ 
\endmatrix } \right)$$
iv) $$\left( {\matrix
   4  \ 
   2  \ 
\endmatrix } \right)\, + \,\left( {\matrix
   4  \ 
   3  \ 
\endmatrix } \right)\, + \,\left( {\matrix
   5  \ 
   4  \ 
\endmatrix } \right)\, + \,\left( {\matrix
   6  \ 
   5  \ 
\endmatrix } \right)\, + \,\left( {\matrix
   7  \ 
   6  \ 
\endmatrix } \right)\, + \,\left( {\matrix
   8  \ 
   7  \ 
\endmatrix } \right)$$ v) $$\left( {\matrix
   4  \ 
   0  \ 
\endmatrix } \right)\, + \,\left( {\matrix
   4  \ 
   1  \ 
\endmatrix } \right)\, + \,\left( {\matrix
   4  \ 
   2  \ 
\endmatrix } \right)\, + \,\left( {\matrix
   4  \ 
   3  \ 
\endmatrix } \right)$$
End of explanation
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