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I have $10$ identical coins that are to be placed at $10$ sites arranged in a ring. Assuming that the coins are placed at random, and by some chance only three coins turned out to be "up-faced". What are the number of possible ways such that none of these three "up-faced" coins are in adjacent sites?

What I tried is that there should be at least one "down-faced" coin in between two up-faced coins. Since, we have $7$ down-faced coins, all I have to find is number of ways to arrange these 7 coins such that there is at least one in between two "down-faced" coins. Note that the answer is $50$, but with my idea I get $38$. Any ideas?

Em.
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titanium
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    Put $n=10$ and $d=3$ into my answer here: http://math.stackexchange.com/questions/154763/combinatorics-selecting-objects-arranged-in-a-circle/154775#154775 –  Mar 28 '16 at 01:19
  • @ByronSchmuland You did not provide a link to your answer. – N. F. Taussig Mar 28 '16 at 01:21
  • @N.F.Taussig Whoops! Thanks for letting me know. –  Mar 28 '16 at 01:22

1 Answers1

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Attach each "upfaced" coin clockwise (say) to a "downfaced" coin, $\boxed{DU}$
They can't now be adjacent.

We now have a total of $7$ objects, $3$ boxes and $4 D's$,
and the boxes can be placed among the lot in $\binom73$ ways.

But by creating objects, we are allowing them only $7$ starting points instead of the actual $10$,

thus # of ways = $\dfrac{10}{7}\dbinom73 = 50$

Added

I know that there are a number of answers to the earlier question, but added this answer as a simple approach that might be of some use.

true blue anil
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