Permutations And Combinations Question 358
Question: 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
