How many $4$-sequences are there on $\{0, \ldots, 9\}$ that start with $0$? I am looking for some clarification to the question "How many $4$-sequences are there on $\{0, \ldots, 9\}$ that start with $0$?" 
A sequence is like a permutation, where order and repetition matters, correct? Then there are $10,000$ total $4$-sequences on $\{0, \ldots, 9\}$ ($0$ through $9999$) AND ONLY $1000$ $4$-sequences on $\{0, \ldots, 9\}$ that start with $0$ ($0-0999$), right? Now this is simple intuition. According to the book, I'm supposed to use $$\frac{n!}{(n-k)!}$$ but I only get $5040$. If I do $$\frac{n!}{k!(n-k)!}$$ I get $216$. Which method do I use?
Thanks!
 A: Don't get bogged down in "which method or formula to use." Think about the problem constructively. Since duplicate numbers are allowed, we have $10$ choices for each of the other three digits, which gives us a total of $\boxed{1000}$ possible sequences.
EDIT: Note that what we just counted was the number of sequences with four terms. If, on the other hand, we wanted to count the number of sets of four numbers selected from $\{1, 2, 3, ..., 9\},$ with repetition allowed, we use casework. If there is one other distinct number, there are $9$ sets. If there are two other distinct numbers, there are $9 \times 8 = 72$ sets. If there are three other distinct numbers, there are $\dbinom{9}{3} = 84$ sets. The total number of sets is $9 + 72 + 84 = \boxed{165}.$
A: Your intuition is correct.  There are $1000$ sequences of length $4$ on the set $\{0, 1, 2, \ldots, 9\}$ that start with $0$.  We have one choice for the first digit, namely 0, and ten choices for each of the three remaining digits.  Thus, by the Multiplication Principle, the number of permissible sequences of length $4$ is $$1 \cdot 10 \cdot 10 \cdot 10 = 1000$$
When you used the formula 
$$\frac{n!}{(n - k)!}$$
with $n = 10$ and $k = 4$ to calculate the number of sequences, you were calculating the number of sequences of length $4$ on the set $\{0, 1, 2, 3, \ldots, 9\}$ in which the numbers are used without repetition, including those that do not start with $0$.  It does not make sense to use the formula in this context since digits are allowed to be repeated.  Had you been asked to find the number of sequences of length $4$ that can be formed from the set $\{0, 1, 2, \ldots, 9\}$ without repetition, you should have obtained 
$$1 \cdot \frac{9!}{(9 - 3)!} = 1 \cdot \frac{9!}{6!} = 1 \cdot 9 \cdot 8 \cdot 7$$ 
since $0$ must be placed in the first position, leaving nine choices for the second position (excluding $0$), eight choices for the third position (excluding the digits in the first two positions), and seven choices for the fourth position (excluding the digits in the first three positions).
When you used the formula 
$$\binom{n}{k} = \frac{n!}{k!(n - k)!}$$
with $n = 10$ and $k = 4$, you were calculating the number of subsets of size $4$ that can be formed from the set $\{0, 1, 2, \ldots, 9\}$, including those that do not include $0$ (of which there are actually $210$).  It does not make sense to use that formula in this context since sequences are ordered and digits are allowed to be repeated.  Had you been asked to find the number of subsets of size $4$ from the set $\{0, 1, 2, \ldots, 9\}$ that include $0$, you should have obtained 
$$1 \cdot \binom{9}{3} = \binom{9}{3}$$
since $0$ must be included and three of the other $9$ elements of the set must be selected. 
