Simple expression
Author: mathforces
Problem has been solved: 84 times
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Suppose that $\frac{C_{54}^{23}+6C_{54}^{24}+15C_{54}^{25}+15C_{54}^{27}+6C_{54}^{28}+C_{54}^{29}-C_{60}^{29}}{C_{54}^{26}}=x$. What is the value of $2020+x^2?$
Note: $C_{n}^{m}=\frac{n!}{m!(n-m)!}$
Пусть $\frac{C_{54}^{23}+6C_{54}^{24}+15C_{54}^{25}+15C_{54}^{27}+6C_{54}^{28}+C_{54}^{29}-C_{60}^{29}}{C_{54}^{26}}=x$. Чему равно $2020+x^2?$
Примечание: $C_{n}^{m}=\frac{n!}{m!(n-m)!}$
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