Calculating Number of Sequences
The problem asks for the number of distinct sequences of length 10 that can be formed using the digits 0, 1, and 2, with the specific conditions that the sequence must contain exactly five 1s and exactly three 2s.
Determining Sequence Composition
A sequence has a total length of 10 terms.
- Number of 1s = 5
- Number of 2s = 3
- Since the terms can only be 0, 1, or 2, the remaining terms must be 0s.
- Number of 0s = Total length - (Number of 1s + Number of 2s) = 10 - (5 + 3) = 10 - 8 = 2.
Therefore, we need to find the number of distinct arrangements of these 10 terms (five 1s, three 2s, and two 0s).
Applying Combinatorics for Sequences
This is a problem of permutations with repetitions, or alternatively, can be solved by selecting positions for each type of digit. We have 10 positions to fill.
- Choose positions for the five 1s: Out of the 10 available positions, we need to choose 5 for the digit 1. The number of ways to do this is given by the combination formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$. Number of ways = $\binom{10}{5} = \frac{10!}{5!(10-5)!} = \frac{10!}{5!5!}$. Calculation: $\binom{10}{5} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252$.
- Choose positions for the three 2s: After placing the five 1s, there are $10 - 5 = 5$ positions remaining. We need to choose 3 of these remaining positions for the digit 2. Number of ways = $\binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5!}{3!2!}$. Calculation: $\binom{5}{3} = \frac{5 \times 4}{2 \times 1} = 10$.
- Place the two 0s: The remaining $5 - 3 = 2$ positions must be filled with the digit 0. There is only one way to do this. Number of ways = $\binom{2}{2} = \frac{2!}{2!(2-2)!} = \frac{2!}{2!0!} = 1$.
Calculating Total Number of Sequences
To find the total number of distinct sequences, we multiply the number of ways to place each digit type:
Total sequences = (Ways to place 1s) $\times$ (Ways to place 2s) $\times$ (Ways to place 0s)
Total sequences = $\binom{10}{5} \times \binom{5}{3} \times \binom{2}{2} = 252 \times 10 \times 1 = 2520$.
Alternatively, we can use the multinomial coefficient formula directly: Total sequences = $\binom{10}{5, 3, 2} = \frac{10!}{5!3!2!} = \frac{3628800}{(120)(6)(2)} = \frac{3628800}{1440} = 2520$.
Thus, there are 2520 possible sequences that meet the given criteria.
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