Question 276012: Ten identically shaped beads are randomly strung along a straight wire. 1f 3 of them are red, 2 are blue, and 5 are green, find the probability the the beads are arranged so that from left to right the beads are red, then blue, then green.
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! Ten identically shaped beads are randomly strung along a straight wire. 1f 3 of them are red, 2 are blue, and 5 are green, find the probability the the beads are arranged so that from left to right the beads are red, then blue, then green.
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If you mean all the red followed by all the blue followed by all the green
there is only one such arrangement.
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Since all the red are indistinguishable (same for blue and for green) there
are 10!/(3!*2!*5!) possible arrangements.
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The probability of the desired arrangement is 1/(10!/(3!*2!*5!)
= (3!*2!*5!)/10!
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Cheers,
Stan H.
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