No. of ways in which 2n white and 2n black balls can be arranged such that no consecutive n white balls are together

Question

The number of ways in which $2n$ white and $2n$ black balls can be arranged such that no consecutive $n$ white balls are together, is

• A. ${}^{2n+1}C_2 + {}^{4n}C_{2n}$
• B. ${}^{2n+1}C_2 - {}^{2n+1}C_1. {}^{3n}C_{n} (-1)^n + {}^{4n}C_{2n}$
• C. ${}^{2n+1}C_2+ (-1)^n. {}^{2n+1}C_1 .  {}^{3n}C_n + {}^{4n}C_{2n}$
• D. ${}^{2n+1}C_2+ (-1)^n. {}^{3n}C_n . {}^{2n+1}C_1$

Comments

options are correct?

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Yes,options are correct.I checked it.

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option b is given as answer.

Answer 1

Without any restriction we have a permutation of $4n$ objects where $2n$ each are identical. So, ${}^{4n}C_{2n}$.

Now we have to subtract the cases where $n$ white balls are consecutive. We can choose the starting position for these $n$ balls in $3n+1$ ways. Remaining $3n$ balls would give ${}^{3n}C_{n}$. So, our answer should be

$${}^{4n}C_{2n} - (3n+1){}^{3n}C_{n}.$$

But this counts repeatedly the $n$ consecutive case- waiting for correct solution.

Comments

question says for no consecutive n white balls rt? not just consecutive.

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ya,thats why I took the value 0 to n-1 (total n) white balls.(But in that 2n+1 places only you have to place the remaining n white balls) I am saying so because after having completed n positions in 2n+1 places we still have( 2n+1-n= n+1 places remaining to place n items).

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No. of ways of putting $2n$ similar balls to $2n+1$ distinct bins such that no bin contains $n$ or more balls :)