Question

The number of sequences of ten terms, whose terms are either 0 or 1 or 2, that contain exactly five 1’s and exactly three 2’s, is equal to:

• A. 360
• B. 45
• C. 2520
• D. 1820

Hint:

Use multinomial coefficients for arranging repeated elements in a sequence.

Answer 1

The Correct Option is C

This problem asks for the number of distinct sequences of ten terms that can be formed using the numbers {0, 1, 2}, with the specific constraints that the sequence must contain exactly five 1’s and exactly three 2’s.

Concept Used:

This is a problem of permutations with repetitions. We have a set of \( n \) objects to arrange, where some of the objects are identical. If there are \( n_1 \) identical objects of type 1, \( n_2 \) identical objects of type 2, ..., and \( n_k \) identical objects of type k, the total number of distinct arrangements is given by the multinomial coefficient formula:

\[ \text{Number of sequences} = \frac{n!}{n_1! n_2! \cdots n_k!} \]

An alternative way to think about this is by using combinations. We can choose the positions for each type of number sequentially.

Step-by-Step Solution:

• Step 1: Determine the composition of the sequence.

The sequence has a total of \( n = 10 \) terms.

The constraints given are:

• Number of 1’s (\( n_1 \)) = 5
• Number of 2’s (\( n_2 \)) = 3
• Step 2: Calculate the number of distinct arrangements using combinations.

We can think of this as filling 10 empty slots in the sequence.

First, we choose the positions for the five 1’s. Out of 10 available positions, the number of ways to place the five 1’s is given by \( \binom{10}{5} \).

• \[ \binom{10}{5} = \frac{10!}{5!(10-5)!} = \frac{10!}{5!5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 2 \times 9 \times 2 \times 7 = 252 \]

After placing the five 1’s, there are \( 10 - 5 = 5 \) positions remaining.

Next, we choose the positions for the three 2’s from the 5 remaining positions. The number of ways to do this is \( \binom{5}{3} \).

• \[ \binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5!}{3!2!} = \frac{5 \times 4}{2 \times 1} = 10 \]

After placing the three 2’s, there are \( 5 - 3 = 2 \) positions remaining.

Finally, the two 0’s must be placed in these last two positions. The number of ways to do this is \( \binom{2}{2} \).

• \[ \binom{2}{2} = \frac{2!}{2!(2-2)!} = \frac{2!}{2!0!} = 1 \]
• Step 3: Calculate the total number of sequences.

Using the multiplication principle, the total number of distinct sequences is the product of the number of ways for each step:

• \[ \text{Total sequences} = \binom{10}{5} \times \binom{5}{3} \times \binom{2}{2} \]

Final Computation & Result:

Substituting the calculated values:

\[ \text{Total sequences} = 252 \times 10 \times 1 = 2520 \]

Alternatively, using the direct formula for permutations with repetitions:

\[ \frac{10!}{5!3!2!} = \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5!}{5! \times (3 \times 2 \times 1) \times (2 \times 1)} = \frac{10 \times 9 \times 8 \times 7 \times 6}{6 \times 2} = 10 \times 9 \times 4 \times 7 = 2520 \]

Thus, the total number of such sequences is 2520.

Answer 2

We are arranging 5 ones, 3 twos, and 2 zeros in 10 positions: \[ \text{Number of sequences} = \frac{10!}{5!3!2!} = 2520 \]