\[^{14}{{C}_{4}}+\sum\limits_{j=1}^{4}{^{18-j}{{C}_{3}}}\] is equal to [EAMCET 1991]

\[^{14}{{C}_{4}}+\sum\limits_{j=1}^{4}{^{18-j}{{C}_{3}}}\] is equal to

1.	\[^{14}{{C}_{4}}+\sum\limits_{j=1}^{4}{^{18-j}{{C}_{3}}}\] is equal to [EAMCET 1991]
A.	\[^{18}{{C}_{3}}\]
B.	\[^{18}{{C}_{4}}\]
C.	\[^{14}{{C}_{7}}\]
D.	None of these
Answer» C. \[^{14}{{C}_{7}}\]

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\[^{14}{{C}_{4}}+\sum\limits_{j=1}^{4}{^{18-j}{{C}_{3}}}\] is equal to [EAMCET 1991]

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