Numbers of strings that begin with consonant Suppose you have the letters $[A, B, C, D, D, F]$. Find the total number of strings that begin with a consonant.
I was thinking there are $5$ consonants to choose from for the first position: $[B, C, D, D, F]$ and that would leave $5$ other characters to permute:
$$\binom{5}{1}\cdot5!$$
But I think the duplicate $D$ introduces a double count, for example, the string $DDBCFA$ would be counted twice. I'm not sure how to avoid overcounting.
 A: We may solve this question faster using complementary counting, i.e. subtracting the number of undesirable cases from the number of total cases. The only undesirable case is when the string begins with A, and we may order the consonants however we like which gives $\frac{5!}{2!}=60$. We divide by $2!$ since there are $2$ D's and thus $2!=2$ ways to order them. I believe this also answers your question on how to deal with overcounting.
Now, we calculate the total number of cases which is $\frac{6!}{2!}=360$ by a similar token, and subtract to get $300$. Notice: $\binom51\cdot5!/2=300$ as well.
A: The simplest way may be to bring probability as an aid, thus
thus P(start with consonant]$\times$ [Total ways]$\,=\frac56\times\frac{6!}{2} = 300$
A: This answer assumes that the OP (i.e. original poster) is forming strings of exactly $6$ characters, where each character is used exactly once.
To avoid over-counting the two D's, my approach is to split the enumerations into two cases:

*

*first character is a D


*first character is not a D.
If the first character is a D, then you have $5$ distinct remaining characters to permute, which can be done in $5!$ ways.
There are $3$ possible other consonants that can be the first character, if the first character is not a D.
Then, the computation of $5!$ is the same, except that you (then) need the over-counting adjustment factor of $\dfrac{1}{2!}$.
So, the final computation is
$$5! + \left[ ~3 \times 5! \times \frac{1}{2} ~\right]$$
$$= 5! \times \left[1 + \frac{3}{2}\right] = 120 \times \frac{5}{2} = 300.$$
A: Start with a $B = \frac{5!}{2} = 60$.
Start with a $C = \frac{5!}{2} = 60$.
Start with a $D = 5! = 120$.
Start with a $F = \frac{5!}{2} = 60$.
Total $300$.
