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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

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

Grade:12th pass

1 Answers

mycroft holmes
272 Points
7 years ago
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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