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Grade 12th passAlgebra

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

Anuj

9 Years agoGrade 12th pass

1 Answer

mycroft holmes

9 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

$\binom {n+m}{m} = \binom {n+m+1}{m}-\binom {n+m}{m-1}$

$\binom {n+m-1}{m-1} = \binom {n+m}{m-1}-\binom {n+m-1}{m-2}$

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$\binom {n+1}{1} = \binom {n+2}{1}-\binom {n+1}{0}$

$\binom {n}{0} = \binom {n+1}{0}$

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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