Problem: We construct a necklace using 7 green and 9 red beads. How many different necklaces can we construct? Necklaces that can be rotated to each other are considered to be the same.
My answer: $\frac{1}{16}\binom{16}{7, 9}$ $= 715$. There are $\binom{16}{7, 9}$ many ways to linearly label $7$ bead positions green and $9$ bead positions red. But that over counts by a factor of $16$ since there are $16$ rotations that give the same necklace but different linear arrangement.
Textbook answer: $\frac{1}{18}$ $18\choose 7$ $= 1768$.
Question: How did they figure $18$ positions for beads on the necklace?
Update: As Brian Moehring has pointed out in the comments I really should also show that the 16 rotations are counted by $16\choose 7$ for my answer/argument to be complete.
Suppose the seven green beads occupy seven of the 16 positions on the necklace, and suppose further that the distance between each of the green beads is $s$ for some $s$ strictly between $0$ and $16$, so that any green bead is $s+1$ positions away from the next. If this were possible, then clearly a rotation/shift by $s+1$ positions does not give a unique linear arrangement and so dividing $16\choose 7$ by 16 would not give an accurate count of the necklaces.
So two questions we need to now answer. One, is such a situation possible in our case - that the distance between each of the green beads is $s$ for some $s$ strictly between $0$ and $16$? And two, is there any other type of spacing between the green beads that would also lead to a rotation or shift of less than 16 positions giving the same linear arrangement?
For the first question we argue by contradiction. Suppose the distance between each of the green beads is $s$ for some $s$ strictly between $0$ and $16$. Since there are $7$ green beads there are $7s$ red beads. Hence, $$ 16 = 7 + 7s = 7(1 + s) $$ which is a contradiction. Therefore, the distance between each of the green beads cannot all be the same.
For the second question we get more visual, I guess. Number the green beads $g_1, g_2, \ldots, g_7$. Let $s_1$ be the distance between $g_1$ and $g_2$, $s_2$ be the distance between $g_2$ and $g_3$, and so on, where some of the $s_i$ could be $0$ and there is an $i$ and $j$ so that $s_i \neq s_j$. Lined up we have $$ g_1\, s_1\, g_2\, s_2\, g_3\, s_3\, g_4\, s_4\, g_5\, s_5\, g_6\, s_6\, g_7\, s_7 $$ WLOG suppose $s_1 \neq s_2$. Then shifting by any amount to put $g_1$ into any other $g_i$ position gives us six possibilities. If indeed we did get the same linear arrangement in one of these possibilities, then we arrive at a contradiction. For if $g_1$ goes to the $g_3$ position, say, then we have $$ g_6\, s_6\, g_7\, s_7\, g_1\, s_1\, g_2\, s_2\, g_3\, s_3\, g_4\, s_4\, g_5\, s_5 $$ which means that $s_1 = s_6 = s_4 = s_2$. Hence, the only way for this to happen is if all of the $s_i$ are the same.