Permutations And Combinations Question 99

Question: $ \sum\limits _{r=0}^{m}{^{n+r}C _{n}=} $ [Pb. CET 2003]

Options:

A) $ ^{n+m+1}{C _{n+1}} $

B) $ ^{n+m+2}C _{n} $

C) $ ^{n+m+3}{C _{n-1}} $

D) None of these

Show Answer

Answer:

Correct Answer: A

Solution:

  • Since $ ^{n}C _{r}{{=}^{n}}{C _{n-r}} $ and $ ^{n}{C _{r-1}}{{+}^{n}}C _{r}{{=}^{n+1}}C _{r} $ we have $ \sum\limits _{r=0}^{m}{^{n+r}C _{n}}=\sum\limits _{r=0}^{m}{^{n+r}C _{r}}{{=}^{n}}C_0{{+}^{n+1}}C_1{{+}^{n+2}}C_2+……{{+}^{n+m}}C _{m} $ $ =[1+(n+1)]{{+}^{n+2}}C_2{{+}^{n+3}}C_3+……..{{+}^{n+m}}C _{m} $ $ {{=}^{n+m+1}}{C _{n+1}} $ , $ [\because {{\ }^{n}}C _{r}{{=}^{n}}{C _{n-r}}] $ .

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