A) \[{{\,}^{47}}{{C}_{6}}\]
B) \[{{\,}^{52}}{{C}_{5}}\]
C) \[{{\,}^{52}}{{C}_{4}}\]
D) none of these
Correct Answer: C
Solution :
\[{{\,}^{47}}{{C}_{4}}+\sum\limits_{r=1}^{5}{{{\,}^{52-r}}{{C}_{3}}}\] \[={{\,}^{51}}{{C}_{4}}{{+}^{50}}{{C}_{3}}{{+}^{49}}{{C}_{3}}{{+}^{47}}{{C}_{3}}{{+}^{47}}{{C}_{4}}\] \[={{\,}^{51}}{{C}_{3}}{{+}^{50}}{{C}_{3}}{{+}^{49}}{{C}_{3}}{{+}^{48}}{{C}_{3}}{{+}^{48}}{{C}_{4}}\] \[={{\,}^{51}}{{C}_{3}}{{+}^{50}}{{C}_{3}}{{+}^{49}}{{C}_{3}}{{+}^{49}}{{C}_{4}}\] \[={{\,}^{51}}{{C}_{3}}{{+}^{50}}{{C}_{3}}{{+}^{50}}{{C}_{4}}\] \[={{\,}^{51}}{{C}_{3}}{{+}^{51}}{{C}_{4}}\] \[={{\,}^{52}}{{C}_{4}}\]
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