Permutation and CombinationMedium
Question
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 -
Options
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
Solution
Case 1 : When all n red balls are taken but no green ball.
Only 1 arrangement is possible.
Case 2 : n red balls and 1 green balls
Number of arrangement =
Case 3 : n red balls and 2 green balls
Number of arrangement =
case m + 1 : n red balls and m green balls
Number of arrangements =
add all cases
1 +
n+1C0 + n+1C1 + n+2C2 + ............+ n+mCm
n+2C1 + n+2C2 + .............+ n+mCm
(∴ nCr + nCr-1 = n+1Cr)
n+3C2 + n+3C3 + ....... + n+mCm
Finally we get the sum as : m+n+1Cm
Only 1 arrangement is possible.
Case 2 : n red balls and 1 green balls
Number of arrangement =
Case 3 : n red balls and 2 green balls
Number of arrangement =
case m + 1 : n red balls and m green balls
Number of arrangements =
add all cases
1 +
n+1C0 + n+1C1 + n+2C2 + ............+ n+mCm
n+2C1 + n+2C2 + .............+ n+mCm
(∴ nCr + nCr-1 = n+1Cr)
n+3C2 + n+3C3 + ....... + n+mCm
Finally we get the sum as : m+n+1Cm
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