The "ring" question asks "how many ways are there to put 6 people in a ring", where two "ways" are identical if each individual has the same left and right neighbours.
The "table" question asks "how many ways are there to put 6 people at a table", where two "ways" are identical if each individual has the same left and right neighbours and the same place at the table.
It is easy to see that since the possible rotations are 6, the answer to the first question will be 6 times the answer to the second question, and indeed 6! = 6 x 5!.
Concerning the "necklace" problem, the difference from the "ring" problem is that you cannot put people upside down without noticing - while you can to so with a necklace, obtaining a pattern which is symmetric. So the number of combination is halved, because two symmetric combinations are considered as one.
There are $4!$ ways of arranging four distinct beads in a row. However, we only care about their relative order when they are arranged in a circle. Therefore, there are four linear arrangements that correspond to each circular arrangement of four distinct beads. For instance, the four linear arrangements at left in the figure below correspond to the single circular arrangement at right in the figure below (imagine taking the bead at the start of a row and moving it to the end of the row to form the next row) since the relative order does not change when we join the ends of a row to form a circle.
Since each circular permutation corresponds to four linear permutations, there are
$$\frac{4!}{4} = 6$$
distinct circular permutations of four beads. Another way to see this is that when we arrange one blue bead, one green bead, one red bead, and one yellow bead in a circle, we can use the blue bead as our reference point. Relative to the blue ball, there are $3! = 6$ ways to arrange the remaining three beads as we proceed clockwise around the circle.
Let's consider what happens when we do the same thing for two black beads and two white beads. Clearly, there are only two possible circular arrangements since either beads of the same color (shade?) are adjacent or they alternate.
As you realized, there are $\binom{4}{2} = 6$ possible linear arrangements of two black beads and two white beads.
Imagine placing the beads on a string. There are four places we can cut the string between successive beads in order to form a linear arrangement. If we do this for the case where the two black beads are adjacent and the two white beads are adjacent, we get four distinct linear arrangements: bbww, bwwb, wbbw, wwbb. If we do this for the case in which the black and white beads alternate, we again get four linear arrangements, only two of which are distinguishable: bwbw, wbwb, bwbw, wbwb. That is why there are six distinguishable linear arrangements rather than eight.
There are two ways we could account for this:
- If we divide the $\binom{4}{2}$ distinguishable linear arrangements by $4$, we do not get an integer. This is because dividing by $4$ counts the four linear arrangements corresponding to the case in which the two black beads are adjacent and the two white beads are adjacent once, but only counts the two distinguishable linear arrangements in which the black and white beads alternate $2/4 = 1/2$ times. We want to count that circular arrangement once, so we must add $1/2$ to the total, which gives $$\frac{1}{4}\binom{4}{2} + \frac{1}{2} = 2$$
as expected.
- There are four places we could cut the string to form a linear arrangement in each case, giving a total of $\binom{4}{2} + 2 = 8$ ways we could cut the strings of our two necklaces. We add two to the number of distinguishable linear arrangements to account for the two repetitions of the linear arrangements bwbw and wbwb in the case when the black and white beads alternate. With that adjustment, we can now divide by $4$ to obtain the number of circular permutations, which is $$\frac{\binom{4}{2} + 2}{4} = 2$$
For the case with two white beads, one black bead, and one red bead, there are
$$\binom{4}{2}2! = 12$$
possible linear arrangements since we must choose two of the four positions for the white beads, then arrange the black bead and red bead in the remaining two positions. Since each circular arrangement of these beads corresponds to four linear arrangements of the beads, as shown in the figure below, there are
$$\frac{12}{4} = 3$$
distinct circular arrangements.
As for the total, we have to choose the color(s) of the beads in the necklace and multiply by the number of possible arrangements for each choice of colors.
Four beads of one color are used: There are three ways to choose the color (black, white, or red) and one possible arrangement of four beads of that color, giving $3 \cdot 1$ such arrangements.
Three balls of one color and one ball of a different color are used: There are three ways to select the color which is used three times and two ways to select the color that will be used once from the remaining two colors. Hence, there are $3 \cdot 2 = 6$ ways of selecting the colors. For each choice, there is one possible arrangement of three beads of one color and one bead of another color in a circle. Hence, there are $6 \cdot 1$ such arrangements.
Two balls each of two colors are used: There are $\binom{3}{2} = 3$ ways of selecting the two colors to be used and two ways to arrange two beads of each color in a circle. Hence, there are $3 \cdot 2 = 6$ such arrangements.
Two balls of one color and one ball each of the other two colors: There are three ways of selecting the color that will be used twice. For each such choice, there are three ways of arranging two beads of one color and one bead each of the other two colors in a circle. Hence, there are $3 \cdot 3 = 9$ such arrangements.
Total: Since the four cases above are mutually exclusive and exhaustive, the number of necklaces with four beads we can form given white, black, and red beads is
$$3 \cdot 1 + 6 \cdot 1 + 3 \cdot 2 + 3 \cdot 3 = 24$$
as claimed.
Best Answer
An explanation for this case (can be extended to the complete case): 9 flowers in all, 3 of them white and $9 - 3 =6$ nowhite. Thus, there are $\binom{9}{3}$ way to arrange this. But you also have to arrange the 6 nonwhites, 4 lavender and 2 yellow, in $\binom{6}{2}$ ways. In all:
$\begin{equation*} \binom{9}{3} \cdot \binom{6}{2} = \frac{9!}{3! 6!} \cdot \frac{6!}{2! 4!} = \frac{9!}{3! 2! 4!} = \binom{9}{3, 2, 4} \end{equation*}$