As you noted, "arrange ... in a row" indicates that order does matter, i.e., we are counting permutations. The reasoning you used in parts (a) and (b) is correct. For part (c), one way to count the permutations is
- $2$ ways to choose exactly one of the bride and groom,
- $6$ ways to choose their position,
- $8\cdot7\cdot6\cdot5\cdot4=6720$ ways to choose and arrange the other five people.
Thus, there are $80640=(2\cdot6)\cdot8\cdot7\cdot6\cdot5\cdot4$ permutations in this case.
Well, the number of ways you can arrange $a,a,a,b,b$ amounts to the number of ways to choose the locations for the three $a$s, since the $b$s will just go in the other two spots. But choosing locations for the $a$s is simply choosing a $3$-element subset of the set of all $5$ locations, of which there are $C[5,3],$ as you noticed.
Let's number our spots--$\{1,2,3,4,5\}.$ Then if we went with the arrangement $aabab,$ this would correspond to the subset $\{1,2,4\},$ for example. On the other hand, the subset $\{2,3,5\}$ would correspond to the arrangement $baaba.$
As you already observed, this only works with two types of objects to permute, since the locations of the first type of object tell the whole story. Otherwise, we can analogously use multinomial coefficients. (Note that the article uses "permutations with repetition" differently than you do.)
Best Answer
For non-negative integers $n$ and $k$ the binomial coefficient $\binom{n}k$ is the number of $k$-sized subsets of a set of $n$ things: itβs the number of different combinations of $k$ of the $n$ elements. The members of a $k$-sized set can be listed in $k!$ different orders, so there are altogether $k!\binom{n}k$ permutations of $k$-sized subsets of the original set of $n$ things, $k!$ orderings of each of $\binom{n}k$ sets.
But $\dbinom{n}k=\dfrac{n!}{k!(n-k)!}$, so $$k!\binom{n}k=k!\cdot\frac{n!}{k!(n-k)!}=\frac{n!}{(n-k)!}=n(n-1)(n-2)\dots(n-k+1)\;,$$
and you often see the formula for the number of permutations of $k$-sized subsets of an $n$-set expressed in one of these last two ways instead of as $k!\binom{n}k$. In some ways these are simpler than the form $k!\binom{n}k$. However, because the binomial coefficients turn out to be rather easy to work with, and because many relationships involving them are known, itβs often convenient to write the number of permutations as $k!\binom{n}k$ in order to take advantage of those known relationships.