A chain needs to join the two clasps of a robe. The chain will contain links of eight different metallic elements. There are to be nine links. Eight are circular in shape with two of silver and one each of gold, iron, cobalt, copper, zinc and nickel. The ninth link in the shape of a figure of eight is made of chromium. The two silver links must not connect together. In how many different ways could we assemble the chain? (Hint. Consider how you put an asymmetric ring on your finger or link your forefingers and thumbs).
I attempted this question using the problem involving beads in a string,taking away the number of permutations when the gold links are connected from the total.
$$\frac{9!}{2} - \frac{8!}{2}$$
My problem is, how would you consider the figure of $8$ link into this? The hints are not really helpful...
The reason the figure of eight link matters is that the chain can be turned over.
First, we arrange the links in a row. Set aside the two silver links for now. Arrange the remaining seven distinct links in a row without binding them together. This creates eight spaces in which we can place the silver links, six between successive links and two at the ends of the row. $$\square L_1 \square L_2 \square L_3 \square L_4 \square L_5 \square L_6 \square L_7 \square$$ To separate the silver links, we must choose two of these eight spaces in which to place a silver link. Now bind the links together. The number of distinguishable ways the nine links can be arranged in a row to form a chain is $$7!\binom{8}{2}$$ We now attach the clasps to the ends of the row in order to form a belt.
Notice that $$7!\binom{8}{2} = \frac{9!}{2!} - 8!$$ To see why you should have subtracted $8!$, place the adjacent silver links in a box. Then you have eight distinct objects to arrange, the box and the other seven links. The silver links can be arranged within the box in only one distinguishable way.
Notice that if we turn the belt over, the chromium figure of eight will be on the opposite hip of the wearer unless it is in the middle of the chain. We have counted the arrangements in the figure of eight is in the middle twice, once when we arrange the links in a row and once when we arrange them in a row in the opposite order, so we must subtract them from the total.
Again, set aside the silver links. Place the figure of eight in the middle position. Arrange the remaining six distinct links in the remaining six positions. Now, as before, we have eight spaces in which we can place the two silver links. Hence, there are $$6!\binom{8}{2}$$
ways to arrange the links so that the figure of eight is in the middle.
Consequently, the number of distinguishable belts we can form is $$7!\binom{8}{2} - 6!\binom{8}{2}$$