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If $\sum\limits_{i = 0}^4 {^{4 + 1}} {C_i} + \sum\limits_{j = 6}^9 {^{3 + j}} {C_j} = {\,^x}{C_y}$ ($x$ is prime number), then which one of the following is incorrect
Minimum value of $(x - y)$ is $4$
Minimum value of $(x + y)$ is $17$
$(x - y)$ and $(x + y)$ will always be co-prime numbers.
$(x - y)$ is always smaller than $(x + y)$
Solution
$\left( {{{\mkern 1mu} ^4}{C_0} + {{\mkern 1mu} ^5}{C_1} + {{\mkern 1mu} ^6}{C_2} + {{\mkern 1mu} ^7}{C_3} + {{\mkern 1mu} ^8}{C_4}} \right)$
$ + {\mkern 1mu} \left( {^9{C_6} + {{\mkern 1mu} ^{10}}{C_7} + {{\mkern 1mu} ^{11}}{C_8} + {{\mkern 1mu} ^{12}}{C_9}} \right)$
$\left. { = ({\,^5}{C_0} + {\,^5}{C_1} + {\,^6}{C_2} + {\,^7}{C_3} + {\,^8}{C_4}} \right)$
$ + {\mkern 1mu} \left( {^9{C_6} + {{\mkern 1mu} ^{10}}{C_7} + {{\mkern 1mu} ^{11}}{C_8} + {{\mkern 1mu} ^{12}}{C_9}} \right)$
$ = {\,^9}{C_4} + \,\left( {^9{C_6} + {\,^{10}}{C_7} + {\,^{11}}{C_8} + {\,^{12}}{C_9}} \right)$
$ = {\,^9}{C_5} + {\,^9}{C_6} + {\,^{10}}{C_7} + {\,^{11}}{C_8} + {\,^{12}}{C_9} = {\,^{13}}{C_9}{\rm{ \,\,\,or}}\,\,\,{{\rm{ }}^{13}}{C_4}$
