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Each of the coins in a collection is distinct, and is either

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Each of the coins in a collection is distinct, and is either [#permalink]  17 Aug 2004, 21:02
Each of the coins in a collection is distinct, and is either silver or gold. In, how many different ways could all the coins be displayed in a row, IF NO 2 COINS OF THE SAME COLOR COULD BE ADJACENT?

1. The display contains an equal number of gold and silver coins.

2. If only the silver coins were displayed, 5040 different arrangements of the silver coins would be possible.

a. Plz explain clearly.
b. what if the provision in CAPS above (i.e., the no-two-coins-of-the-same-color-together rule) is taken out? how does that affect the probability. I had an instructor tell me it wouldn't affect the overall number of permutations, if one were to determine this. Is that right? It doesn't seem so. how would this solution differ mathematically?

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1. Insufficient because we don't have any info on number of either silver or gold coins.

2. Insufficient because we don't have any info on the number of gold coins

Together it is sufficient - we have the needed info to solve this arrangement question. We will not solve intentionally.

1. 5040 is 7!. Which means we have 7 silver coins and hence 7 Gold coins (as per stem 1).
2. So the no. of ways to arrange would be 14! if there are NO conditions whatsoever.
3. However, if it is as stated in the problem under discussion, we would then we have 7! * 7! * 2 which is lot lesser than 14!. Otherwise, it is simple to state that 14! includes this arrangement of alternate gold and silver.

4. Adll. Info: If there are n items, 'r' of which are identical (say, two gold coins that look exactly the same or two silver coins that look exactly the same) then the number of ways to arrange them would be n!/r!. Note that, in the stated problem the gold and silver coins are 'distinct'.
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