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`        There are n identical red balls & m identical green balls. The number of different linear arrangements consisting of "n red balls but not necessarily all the green balls" is xCy then(A) x = m + n, y = m                              (B) x = m + n + 1, y = m(C) x = m + n + 1, y = m + 1                 (D) x = m + n, y = n`
2 years ago

mycroft holmes
272 Points
```							Suppose there are k green balls, then there are (n+k) balls to be arranged. So if we choose the k places into which the green balls go, the red balls go into the remaining places. Hence the number of arrangements = $\binom {n+k}{k}$ Hence we have to sum $\binom {n}{0}+\binom {n+1}{1}+\binom {n+2}{2}+\cdots +\binom {n+m}{m}$ Using the identity$\binom {n}{r} = \binom {n-1}{r}+\binom {n-1}{r-1}$ we can write$\binom {n+m}{m} = \binom {n+m+1}{m}-\binom {n+m}{m-1}$ $\binom {n+m-1}{m-1} = \binom {n+m}{m-1}-\binom {n+m-1}{m-2}$...$\binom {n+1}{1} = \binom {n+2}{1}-\binom {n+1}{0}$$\binom {n}{0} = \binom {n+1}{0}$ Adding this telescopic sum, we obtain the given sum in closed form as $\boxed{\binom {n+m+1}{m} }$ Thus we have x = n+m+1, y =m
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2 years ago
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• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions