If the two [tex]b[/tex]s are considered distinct, then the number of such words is
[tex]7!=5040[/tex]
If the two [tex]b[/tex]s are indistinct,
[tex]\dfrac{7!}{2!}=\dfrac{5040}2=2520[/tex]
To clarify, "quibble" has 7 possible character positions, and as we draw 1 letter from the pool [tex]\{q, u, i, b, b, l, e\}[/tex], we have 1 less letter to choose from for the next character position. So there are
[tex]7\times6\times5\times4\times3\times2\times1=7!=5040[/tex]
possible words.
If, however, we consider the two [tex]b[/tex]s to be indistinct, so that, for example, [tex]quib_1b_2le[/tex] and [tex]quib_2b_1le[/tex] both count as the same word, then we have to divide the previous total by the number of ways we can rearrange the [tex]b[/tex]s, which is [tex]2\times1=2!=2[/tex].
