Arrangements with Restrictions

1. Restriction Type 1: Items Must Be Together (The Block Method)

Let's start with a common restriction: a group of items must always be arranged together. For example, arranging books on a shelf so that all the fiction books are grouped together.

The strategy here is intuitive: if a set of items must always be together, we can mentally "glue" them into a single block. Then, we perform two main steps:

1. Arrange this new "block" along with the other individual items.
2. Arrange the items inside the block itself.

Let's look at a simple visual example.

To solve this, we follow the logic:

• Step 1: Arrange the blocks. We have two blocks: the Number Block and the Letter Block. We can arrange these two blocks in $2!=2$ ways: (Number Block, Letter Block) or (Letter Block, Number Block).
• Step 2: Arrange items within the blocks.
  • The numbers (1, 2, 3) can be arranged within their block in $3!=6$ ways.
  • The letters (A, B) can be arranged within their block in $2!=2$ ways.
• Step 3: Combine the results. Using the Fundamental Counting Principle, the total number of arrangements is $2!\times 3!\times 2!=2\times 6\times 2=24$.

This powerful technique is often called the Block Method or String Method. To see it applied to more examples, including cases with repeated items, let's turn to a resource.

As you saw, the general procedure is:

1. Identify the items that must stay together and treat them as a single conceptual block.
2. Calculate the number of ways to arrange this block along with the remaining individual items. Remember to account for any identical items here.
3. Calculate the number of ways the items can be arranged within the block. Again, account for any identical items if they exist inside the block.
4. Multiply the results from steps 2 and 3 to get the total number of arrangements.

2. Restriction Type 2: Items Must Be Separated (The Gap Method)

Now, let's tackle the opposite problem: arranging items such that certain ones are never together. For example, seating boys and girls in a row so that no two girls sit next to each other.

Trying to count this directly is difficult. A much better approach is the Gap Method. The logic is as follows:

1. First, place all the items that have no restrictions on them.
2. This arrangement creates "gaps" between the items (and also at the two ends).
3. Then, place the restricted items into these gaps. Since each gap can hold at most one item, this ensures they will never be together.

Let's illustrate with an example: Arrange 3 boys and 2 girls in a row so that the girls are never together.

• Step 1: Arrange the unrestricted items. The 3 boys (B) have no restrictions. Let's arrange them first.
  $\_B_1\_B_2\_B_3\_$
  The boys can be arranged in $3!=6$ ways.
• Step 2: Identify the gaps. The arrangement of 3 boys creates 4 possible gaps for the girls to occupy (marked by underscores).
• Step 3: Place the restricted items in the gaps. We need to place 2 girls into these 4 available gaps. This is a permutation problem (placing Girl 1 in gap 1 and Girl 2 in gap 2 is different from the reverse). The number of ways to arrange 2 girls in 4 gaps is ${}^4{P}_2=4\times 3=12$.
• Step 4: Combine the results. The total number of arrangements is (ways to arrange boys) $\times$ (ways to place girls in gaps) = $3!{\times }^4{P}_2=6\times 12=72$.

This method is systematic and very reliable. Let's see some more examples from the same resource.

An Alternative for "Never Together"

For some simpler "never together" problems, you can also use an indirect method based on what we learned in the first section:

Total Arrangements - Arrangements Where They Are Always Together = Arrangements Where They Are Not Together

For our example of 3 boys and 2 girls:

• Total arrangements without any restrictions: $5!=120$.
• Arrangements where girls are always together (using the Block Method):
  • Treat the 2 girls as one block [GG]. We arrange 3 boys and [GG], which is 4 items: $4!$ ways.
  • The girls inside the block can be arranged in $2!$ ways.
  • Total "together" arrangements = $4!\times 2!=24\times 2=48$.
• Arrangements where girls are not together: $120-48=72$.

This method works well, but be careful. It is most reliable when you want to ensure a specific pair of items are not together. The Gap Method is generally more robust, especially for problems that state "no two of a group of three or more items can be together". For the CAT, being proficient with the Gap Method is highly valuable.

Conclusion

Today we added two crucial techniques to your combinatorics toolkit, allowing you to solve complex arrangement problems with confidence.

Key Takeaways:

• "Always Together" Restriction: Use the Block Method. Treat the required items as a single unit, arrange the units, and then arrange the items within the unit.
• "Never Together" Restriction: Use the Gap Method. Arrange the unrestricted items first to create gaps, then place the restricted items into those gaps.
• The key is to transform a problem with restrictions into a sequence of simpler, unrestricted permutation problems that you already know how to solve.

Preview of the Next Lesson:
We've spent two lessons focusing on permutations (arrangements). Now, we will shift our focus back to combinations (selections). In the next lesson, we will learn how to solve selection problems involving 'at least'/'at most' conditions, which are another very common type of question in the CAT.