Permutations And Combinations Question 44
Question: There are $ (n+1) $ white and $ (n+1) $ black balls each set numbered 1 to $ n+1 $ . The number of ways in which the balls can be arranged in a row so that the adjacent balls are of different colours is [EAMCET 1991]
Options:
A) $ (2n+2)\ ! $
B) $ (2n+2)\ !\ \times 2 $
C) $ (n+1)! \times 2 $
D) $ 2{{{(n+1)\ !}}^{2}} $
Correct Answer: D Since the balls are to be arranged in a row so that the adjacent balls are of different colours, therefore we can begin with a white ball or a black ball. If we begin with a white ball, we find that $ (n+1) $ white balls numbered 1 to $ (n+1) $ can be arranged in a row in $ (n+1)\ ! $ ways. Now $ (n+1) $ places are created between $ n $ white balls which can be filled by $ (n+1) $ black balls in $ (n+1)\ ! $ ways. So the total number of arrangements in which adjacent balls are of different colours and first ball is a white ball is $ (n+1)\ !\ \times (n+1)\ !\ ={{[(n+1)\ !]}^{2}} $ . But we can begin with a black ball also. Hence the required number of arrangements is $ 2{{[(n+1)\ !]}^{2}} $ .Show Answer
Answer:
Solution:
BETA
