Counting Arrangements of 0s and 1s with No Consecutive 1s: A Combinatorial Approach

Are you looking for ways to solve a mathematical problem involving the arrangement of numbers without consecutive duplicates? One such problem is determining the number of arrangements of 10 zeros and 5 ones with no two consecutive ones. This article will guide you through the combinatorial approach used to solve this problem. We will explore each step in detail, ensuring clarity and understanding.

Introduction to the Problem

The problem we are addressing is: how many arrangements of ten 0s and five 1s can be made with no consecutive 1s? To tackle this, we will use a classic combinatorial approach involving gaps and the binomial coefficient.

Step 1: Arrange the 0s

To find the desired arrangements, we start by placing the ten 0s. This step creates 11 possible gaps in the sequence:

_ 0 _ 0 _ 0 _ 0 _ 0 _ 0 _ 0 _ 0 _ 0 _ 0 _ 0 _

These gaps are numbered as follows:

One gap before the first 0 One gap between each pair of 0s (nine pairs) One gap after the last 0

In total, there are 11 gaps available for placing the 1s.

Step 2: Place the 1s

Now, we need to insert five 1s into these 11 gaps, ensuring that no two 1s occupy the same gap to prevent consecutive 1s.

Step 3: Choose Gaps for the 1s

The task of choosing 5 gaps out of 11 to place the 1s can be approached using the binomial coefficient. The formula for the binomial coefficient is given by:

nCr frac{n!}{r!(n-r)!}

For our specific case:

11C5 frac{11!}{5!6!} frac{39916800}{120 cdot 720} 462

The multinomial coefficient (11C5) calculates how many ways we can choose 5 gaps from 11 available gaps, ensuring no two 1s are consecutive.

Conclusion

Therefore, the total number of arrangements of ten 0s and five 1s with no consecutive 1s is:

boxed{462}

This combinatorial approach using gaps and the binomial coefficient ensures a systematic and accurate way to solve the problem. Understanding the steps involved can help in solving similar problems involving distinct sequences and arrangements.

Additional Insights

It’s interesting to note that another approach can be used to arrive at the same solution. Let’s consider an alternative method:

Imagine 10 zeros with 11 gaps for the 1s. We need to choose 5 gaps from these 11 for the 1s. Alternatively, we can use the direct multiplication method:

11 ways for the first 1, 10 ways for the second, 9 ways for the third, 8 ways for the fourth, and 7 ways for the fifth.

11 * 10 * 9 * 8 * 7 55,440

However, since we only have 5 ones, we need to factor in the permutations of these 5 ones, which is 5!. Therefore, the total number of unique arrangements using this method is:

frac{55440}{120} 462

This confirms the initial calculation and the combinatorial approach.