In how many ways 10 blue marbles and 5 green marbles can be arranged in a row such that no two green marbles are together ?

Can anyone help me solving this.

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- Nov 13th 2010, 02:07 AMkumaran5555permutaions
In how many ways 10 blue marbles and 5 green marbles can be arranged in a row such that no two green marbles are together ?

Can anyone help me solving this. - Nov 13th 2010, 02:31 AMemakarov
Suppose you have a row of 10 blue marbles. Then you have to insert green marbles in this row. There is a certain number of places, let's call it n, where you can put those 5 green marbles. Therefore, the number of ways is $\displaystyle n\choose 5$.

See also this Wikipedia article about multiset coefficients, which are similar to binomial coefficients. In concerns with breaking the number 10 into 6 summands (by placing 5 green marbles between 10 blue ones) all of which are either positive or nonnegative. In your case, the leftmost and rightmost groups of blue marbles can be empty while other groups must be nonempty, but the idea of the solution is similar.