Dare to arrange them!

Probability Level pending

Calculate the number of ways to arrange n n 1 1 s and n n 2 2 s in a row such that, up to any point in the row, the number of 1 1 's is greater than or equal to the number of 2 2 's.

( 2 n ! ) / ( 2 n ) (2n!)/(2^n) ( 2 n ! ) / ( ( n ! ) ( 2 n ) ) (2n!)/((n!)(2^n) ) ( 2 n ! ) / ( ( n ! ) ( ( n ) ! ) ) (2n!)/((n!)((n)!) ) ( 2 n ! ) / ( ( n ! ) ( ( n + 1 ) ! ) ) (2n!)/((n!)((n+1)!))

Mrigank Krishan by Mrigank Krishan , 20, India

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1 solution

Dion Solang
Feb 2, 2019

For the case of n=2 we have only 2 ways to arrange them. By plugging n=2 to the options, we get that only the second option that equals to 2.

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