Question

There are \[\left( {n + 1} \right)\]white and \[\left( {n + 1} \right)\]black balls each set numbered \[1\] to \[n + 1\]. The number of way which the balls can be arranged in a row so that the adjacent balls are of different colors are
A.\[\left( {2n + 2} \right)!\]
B.\[\left( {2n + 2} \right)!{\text{ }} \times 2\]
C.\[\left( {n + 1} \right)! \times 2\]
D.\[2{\left\{ {\left( {n + 1} \right)!} \right\}^2}\]

Answer 1

Hint: We will take two considerations of the possible arrangement . We can put the white ball first then, black or we can put black ball . We will find the number of ways by using the permutation formula . At last, we will add up the total number of ways .

Complete step-by-step answer:
Given : Number of white balls \[ = n + 1\]
Number of white balls \[ = n + 1\] .
We can will take two cases of the possible arrangement :
Case 1 :
Let the row be initiated by placing the white ball . Therefore number of ways of selecting the white ball \[ = \left( {n + 1} \right)\]
Now , the adjacent place is to filled by black ball , therefore number of ways for placing the black ball are \[ = n + 1\]
Now , the third ball is to be white for that number of balls available are \[ = n\]
Therefore , number of ways for arranging all the white balls are \[ = \left( {n + 1} \right)!\]
Similarly , for black balls number of ways are \[ = \left( {n + 1} \right)!\]
Total number of ways arranging the \[ = n + 1\] white balls and \[ = n + 1\] black balls are \[ = \left( {n + 1} \right)! \times \left( {n + 1} \right)!\]
\[ = {\left\{ {\left( {n + 1} \right)!} \right\}^2}\]
Case 2 :
Let the row be initiated by placing the black ball . Therefore number of ways of selecting the black ball \[ = \left( {n + 1} \right)\]
Now , the adjacent place is to filled by white ball , therefore number of ways for placing the white ball are \[ = n + 1\]
Now , the third ball is to be black for that number of balls available are \[ = n\]
Therefore , number of ways for arranging all the white balls are \[ = \left( {n + 1} \right)!\]
Similarly , for black balls number of ways are \[ = \left( {n + 1} \right)!\]
Total number of ways arranging the \[\left( {n + 1} \right)\] white balls and \[\left( {n + 1} \right)\] black balls are \[ = \left( {n + 1} \right)! \times \left( {n + 1} \right)!\]
\[ = {\left\{ {\left( {n + 1} \right)!} \right\}^2}\]
Now , total number ways are \[ = {\left\{ {\left( {n + 1} \right)!} \right\}^2} + {\left\{ {\left( {n + 1} \right)!} \right\}^2}\]
\[ = 2{\left\{ {\left( {n + 1} \right)!} \right\}^2}\]
Therefore , option (D) is the correct answer .
So, the correct answer is “Option D”.

Note: In the given question it is not given whether to start with which ball , it can either be white or black . If it is given that you have start the row with black ball then , the total number of ways would be only \[{\left\{ {\left( {n + 1} \right)!} \right\}^2}\] , not \[ = 2{\left\{ {\left( {n + 1} \right)!} \right\}^2}\] .