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A necklace is made out of $10$ different red, $20$ different orange and $30$ same yellow pearls. If we know that no two red pearls are next to each other, what is the probability that between each couple of red pearls there is at least $1$ orange and from $2$ to $5$ yellow pearls?

I got stuck when I tried to count the set $A$ = {no two red pearls are next to each other }. My idea was to fix one red and then from all possible circular permutations subtract the permutations where there is one red pearl right next that fixed one, or where some other two reds are together, then analogously with three reds together, etc.

It seems to me that this way I would count a lot more than I should, but I cannot think of any other way.

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    $\begingroup$ That's not how I'd try to count those cases. Figure out how many circular permutations there are of the orange and yellow pearls. For each distinct permutation, figure out how many ways there are to place the red pearls so that they're never adjacent to each other. $\endgroup$ Apr 14, 2019 at 9:06
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    $\begingroup$ I would say if "we know that no two red pearls are next to each other", so the probability of this fact is 1 and it is not necessary to solve it further. $\endgroup$
    – georg
    Apr 14, 2019 at 9:09
  • $\begingroup$ Furthermore, I would say that there are certainly 1 orange and 2 yellow pearls between each pair of red pearls, so only the remaining 10 orange and 10 yellow pearls can be considered. $\endgroup$
    – georg
    Apr 14, 2019 at 9:24

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