Key Concepts and Formulas

• Permutations with Repetition: The number of distinct arrangements of $n$ objects, where there are $n_1$ identical objects of type 1, $n_2$ identical objects of type 2, ..., and $n_k$ identical objects of type k, is given by $\frac{n!}{n_1! \cdot n_2! \cdot \ldots \cdot n_k!}$, where $n_1 + n_2 + \dots + n_k = n$.
• Factorial: $n! = n \times (n-1) \times (n-2) \times \dots \times 2 \times 1$.
• Combination: The number of ways to choose $k$ objects from a set of $n$ distinct objects is given by $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.

Step-by-Step Solution

Step 1: Determine the Number of Each Term Type

We need to find the number of 0s in the sequence. We are given that the sequence has ten terms, five 1s, and three 2s. Since the sequence only contains 0s, 1s, and 2s, the number of 0s can be found as follows:

$\text{Number of 0s} + \text{Number of 1s} + \text{Number of 2s} = \text{Total terms}$ $n_0 + 5 + 3 = 10$ $n_0 = 10 - 5 - 3$ $n_0 = 2$

So, there are two 0s in the sequence. We now have:

• Five 1s
• Three 2s
• Two 0s
• Total: 10 terms

Step 2: Apply the Permutation Formula

We have a sequence of 10 terms with repetitions. We can use the permutations with repetition formula to find the number of distinct sequences:

$\text{Number of sequences} = \frac{n!}{n_1! \cdot n_2! \cdot n_3!}$

Where $n = 10$ (total terms), $n_1 = 5$ (number of 1s), $n_2 = 3$ (number of 2s), and $n_3 = 2$ (number of 0s).

Substituting these values into the formula:

$\text{Number of sequences} = \frac{10!}{5! \cdot 3! \cdot 2!}$

Step 3: Calculate the Value

We now calculate the value of the expression:

$\frac{10!}{5! \cdot 3! \cdot 2!} = \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5!}{5! \times 3 \times 2 \times 1 \times 2 \times 1}$

We can cancel out the $5!$ terms:

$= \frac{10 \times 9 \times 8 \times 7 \times 6}{3 \times 2 \times 1 \times 2 \times 1}$

Simplify the denominator: $3 \times 2 \times 1 = 6$ and $2 \times 1 = 2$. So, the denominator is $6 \times 2 = 12$.

$= \frac{10 \times 9 \times 8 \times 7 \times 6}{12}$

We can cancel out the 6 in the numerator and denominator:

$= \frac{10 \times 9 \times 8 \times 7}{2}$

Further simplification:

$= 10 \times 9 \times 4 \times 7$ $= 90 \times 28$ $= 2520$

Common Mistakes & Tips

• Double-check the sum: Ensure that the number of each element (0s, 1s, and 2s) adds up to the total number of terms (10).
• Simplify before calculating: To avoid large numbers, cancel out common factors in the numerator and denominator before performing the full calculation.
• Understand the formula: Make sure you understand the permutations with repetition formula and when it is applicable.

Summary

We determined that the sequence contains five 1s, three 2s, and two 0s. Using the formula for permutations with repetition, we found the number of distinct sequences to be $\frac{10!}{5! \cdot 3! \cdot 2!} = 2520$.

Final Answer

The final answer is \boxed{2520}, which corresponds to option (B).