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In how many ways 10 identical blue marbles and 5 identical green marbles be arranged in a row so that no 2 green marbles be together?

I know this is star and bars problem but, I am not getting to the answer?

Also, same problem but the restriction with regards to the green marbles is removed.

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3 Answers 3

up vote 2 down vote accepted

Add another blue marble, attach a blue marble to the right of each green marble, arrange the $5$ pairs and $6$ blue marbles, then undo the pairs and remove the right-most blue marble. This generates each admissible arrangement exactly once, so there are $\binom{5+6}5=462$ of them.

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Thank god I have good grasp. That book has the answer wrong then. Even I got the same answer. –  Master Chief Feb 3 '13 at 11:03

Consider taking out all the 10 blue marbles and laying them in a line $$B\space\space B\space\space B\space\space B\space\space B\space\space B\space\space B\space\space B\space\space B\space\space B\space\space $$ Now, you do not want any 2 green marbles to be together. What you can do is that you can think of 11 slots surrounding the blue marbles, here represented by hearts.

$$ \heartsuit \space\space B\space\space \heartsuit \space\space B\space\space \heartsuit \space\space B\space\space \heartsuit \space\space B\space\space \heartsuit \space\space B\space\space \heartsuit \space\space B\space\space \heartsuit \space\space B\space\space \heartsuit \space\space B\space\space \heartsuit \space\space B\space\space \heartsuit \space\space B\space\space \heartsuit \space\space$$

You can select any 5 hearts and put green marbles in. You mentioned all the marbles are identical (at least, for the same coloured ones). Hence, you can just employ $${{11}\choose {6}} = 462$$ and you have your final answer.

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Line up the blue marbles with gaps between them, like this: $$\text{B}\quad\text{B}\quad\text{B}\quad\text{B}\quad\text{B}\quad\text{B}\quad\text{B}\quad\text{B}\quad\text{B}\quad\text{B}\quad.$$

The blue marbles give rise to $11$ "gaps." We are counting the space before the leftmost B, and the space after the rightmost b, as a "gap."

We choose $5$ of these gaps to slip a green marble into. There are $\dbinom{11}{5}$ ways to do the choosing.

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