A) \[^{n+1}{{C}_{r-1}}\]
B) \[^{n+2}{{C}_{r-1}}\]
C) \[^{n+3}{{C}_{r-1}}\]
D) \[^{n+4}{{C}_{r-1}}\]
Correct Answer: C
Solution :
\[^{n+4}{{C}_{r}}-{{(}^{n}}{{C}_{r}}+{{3.}^{n}}{{C}_{r-1}}+{{3.}^{n}}{{C}_{r-2}}{{+}^{n}}{{C}_{r-2}}{{+}^{n}}{{C}_{r-3}})\] \[{{=}^{n+4}}{{C}_{r}}-{{(}^{n}}{{C}_{r}}{{+}^{n}}{{C}_{r-1}})+{{(}^{n}}{{C}_{r-2}}{{+}^{n}}{{C}_{r-2}})\] \[+{{(}^{n}}{{C}_{r-1}}{{+}^{n}}{{C}_{r-2}})+{{(}^{n}}{{C}_{r-2}}{{+}^{n}}{{C}_{r-3}})\] \[{{=}^{n+4}}{{C}_{r}}{{-}^{n+1}}{{C}_{r}}{{+}^{n+1}}{{C}_{r-1}}{{+}^{n+1}}{{C}_{r-1}}{{+}^{n+1}}{{C}_{r-2}}\] \[{{=}^{n+4}}{{C}_{r}}{{-}^{n+2}}{{C}_{r}}{{+}^{n+2}}{{C}_{r-1}}\] \[{{=}^{n+4}}{{C}_{r}}{{-}^{n+3}}{{C}_{r}}{{=}^{n+3}}{{C}_{r-1}}\]You need to login to perform this action.
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