A) \[^{50}{{C}_{4}}\]
B) \[^{50}{{C}_{4}}\]
C) \[^{10}{{C}_{5}}\]
D) \[^{52}{{C}_{6}}\]
Correct Answer: D
Solution :
\[^{47}{{C}_{4}}+\sum\limits_{i=1}^{5}{^{52-i}{{C}_{3}}}\] \[{{=}^{47}}{{C}_{4}}{{+}^{51}}{{C}_{3}}{{+}^{50}}{{C}_{3}}{{+}^{49}}{{C}_{3}}{{+}^{48}}{{C}_{3}}{{+}^{47}}{{C}_{3}}\] \[{{=}^{47}}{{C}_{4}}{{+}^{47}}{{C}_{3}}{{+}^{48}}{{C}_{3}}{{+}^{49}}{{C}_{3}}{{+}^{50}}{{C}_{3}}{{+}^{51}}{{C}_{3}}\] \[{{=}^{48}}{{C}_{4}}{{+}^{48}}{{C}_{3}}{{+}^{49}}{{C}_{3}}{{+}^{50}}{{C}_{3}}{{+}^{51}}{{C}_{3}}\] \[{{=}^{49}}{{C}_{4}}{{+}^{49}}{{C}_{3}}{{+}^{50}}{{C}_{3}}{{+}^{51}}{{C}_{3}}\] \[{{=}^{50}}{{C}_{4}}{{+}^{50}}{{C}_{3}}{{+}^{51}}{{C}_{3}}\] \[{{=}^{51}}{{C}_{4}}{{+}^{51}}{{C}_{3}}\] \[{{=}^{52}}{{C}_{4}}\]
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