Permutations And Combinations Question 358
The expression $ ^{n}C _{r}+{4}^{n}C _{r-1}+{6}^{n}C _{r-2}+{4}^{n}C _{r-3} $ $ +^{n}C _{r-4} $ is equal to
Options:
A) $ ^{n+4}C _{r} $
B) $ {{2.}^{n+4}}{C _{r-1}} $
C) $ {{4.}^{n}}C _{r} $
D) $ {{11.}^{n}}C _{r} $
Correct Answer: A $ ={{(}^{n}}C _{r}{{+}^{n}}{C _{r-1}})+3{{(}^{n}}{C _{r-1}}{{+}^{n}}{C _{r-2}})$ $ +3{{(}^{n}}{C _{r-2}} + {^{n}}{C _{r-3}}) + ({^{n}}{C _{r-3}} + {^{n}}{C _{r-4}}) $ $ ={{n+1}}C _{r}+{{3}}^{n+1}{C _{r-1}}+{{3}}^{n+1}{C _{r-2}}+{{n+1}}{C _{r-3}}$ $ ={{(}^{n+1}}C _{r}+{^{n+1}}C _{r-1}})+2{{(}^{n+1}}C _{r-1}}+ $ $ +{{(}^{n+1}}{C _{r-2}} + {^{n+1}}{C _{r-3}}) $ $ {{=}^{n+2}}C _{r}+2({}^{n+2}C _{r-1}+{}^{n+2}C _{r-2}) $ $ ={{(}^{n+2}}C _{r}{{+}^{n+2}}C _{r-1}})+{{(}^{n+2}}C _{r-1}{{+}^{n+2}}C _{r-2}})$ $ {{=}^{n+3}}C _{r}{{+}^{n+3}}{C _{r-1}}{{=}^{n+4}}C _{r} $Show Answer
Answer:
Solution:
BETA
