In a meeting of 3 representatives from each of 6 different companies, each person shook hands with every person not...

1. Translate the problem requirements: We have 6 companies with 3 representatives each (18 people total). Each person shakes hands with every person from the other 5 companies, but not with anyone from their own company. We need to count total handshakes without double-counting.
2. Calculate handshakes between company pairs: Determine how many handshakes occur between any two specific companies, since representatives from one company shake hands with all representatives from another company.
3. Count total company pairs: Find how many distinct pairs of companies exist, since handshakes only occur between people from different companies.
4. Calculate total handshakes: Multiply handshakes per company pair by the number of company pairs to get the final answer.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we have here in simple terms:

We're at a business meeting with 6 different companies. Each company sent exactly 3 representatives, so we have \(6 \times 3 = 18\) people total.

Now, here's the key rule: each person shakes hands with every person who is NOT from their own company. Think of it like this - if you work for Apple, you'll shake hands with everyone from Google, Microsoft, Amazon, etc., but you won't shake hands with your own Apple colleagues (they're your teammates, after all).

We need to count the total number of handshakes, but we have to be careful not to count the same handshake twice. When John from Apple shakes hands with Sarah from Google, that's the same handshake as when Sarah from Google shakes hands with John from Apple.

Process Skill: TRANSLATE - Converting the business scenario into clear mathematical relationships

2. Calculate handshakes between company pairs

Let's think about what happens between any two companies. Say we pick Apple and Google.

Apple has 3 representatives: \(\mathrm{A_1}\), \(\mathrm{A_2}\), \(\mathrm{A_3}\)
Google has 3 representatives: \(\mathrm{G_1}\), \(\mathrm{G_2}\), \(\mathrm{G_3}\)

Each Apple person shakes hands with each Google person:

• \(\mathrm{A_1}\) shakes hands with \(\mathrm{G_1}\), \(\mathrm{G_2}\), \(\mathrm{G_3}\) (that's 3 handshakes)
• \(\mathrm{A_2}\) shakes hands with \(\mathrm{G_1}\), \(\mathrm{G_2}\), \(\mathrm{G_3}\) (that's 3 more handshakes)
• \(\mathrm{A_3}\) shakes hands with \(\mathrm{G_1}\), \(\mathrm{G_2}\), \(\mathrm{G_3}\) (that's 3 more handshakes)

So between Apple and Google, we have \(3 \times 3 = 9\) handshakes.

This pattern holds for any pair of companies: 3 representatives from the first company times 3 representatives from the second company = 9 handshakes per company pair.

3. Count total company pairs

Now we need to figure out how many different pairs of companies we can form from our 6 companies.

Let's list them out systematically. If our companies are A, B, C, D, E, F, then our pairs are:

• A with B, C, D, E, F (that's 5 pairs)
• B with C, D, E, F (that's 4 more pairs - we already counted A-B)
• C with D, E, F (that's 3 more pairs)
• D with E, F (that's 2 more pairs)
• E with F (that's 1 more pair)

Total pairs: 5 + 4 + 3 + 2 + 1 = 15 pairs of companies.

We can also think of this as choosing 2 companies from 6: \(\mathrm{C}(6,2) = \frac{6!}{2! \times 4!} = \frac{6 \times 5}{2 \times 1} = 15\).

Process Skill: VISUALIZE - Breaking down the pairing systematically prevents double-counting

4. Calculate total handshakes

Now we can put it all together:

• Handshakes between each pair of companies: 9
• Number of company pairs: 15
• Total handshakes: \(9 \times 15 = 135\)

Let's verify this makes sense: Each person shakes hands with 5 companies × 3 people per company = 15 people. With 18 total people, that would be \(18 \times 15 = 270\) total "handshake events." But since each handshake involves two people, we've double-counted, so we divide by 2: \(270 ÷ 2 = 135\). ✓

Final Answer

The total number of handshakes is 135.

Looking at our answer choices, this matches choice B. 135.

Common Faltering Points

Errors while devising the approach

1. Misinterpreting the handshake constraint
Students may misread "each person shook hands with every person not from his or her own company" and think that people DO shake hands with their own company colleagues. This leads to counting ALL possible handshakes among 18 people, resulting in \(\mathrm{C}(18,2) = 153\) handshakes instead of the correct 135.

2. Confusion about what constitutes "one handshake"
Students might not recognize that when Person A shakes hands with Person B, this is the same physical handshake as when Person B shakes hands with Person A. This fundamental misunderstanding about the nature of handshakes can lead to double-counting throughout their approach.

3. Attempting to count individual handshakes rather than using systematic grouping
Some students may try to list out every single handshake individually (Person 1 with Person 4, Person 1 with Person 5, etc.) rather than recognizing the more efficient approach of counting handshakes between company pairs. This makes the problem unnecessarily complex and error-prone.

Errors while executing the approach

1. Incorrect calculation of company pairs
When counting how many pairs of companies exist, students might incorrectly calculate \(6 \times 5 = 30\) instead of \(\mathrm{C}(6,2) = 15\). They fail to recognize that pairing Company A with Company B is the same as pairing Company B with Company A, leading to double-counting of company pairs.

2. Arithmetic errors in the final multiplication
Even with the correct approach, students may make simple computational mistakes when calculating \(9 \times 15\), potentially getting 145 instead of 135, or making errors in the alternative verification method of \((18 \times 15) ÷ 2\).

3. Double-counting handshakes in the total calculation
Students who correctly identify that each person shakes hands with 15 others (5 companies × 3 people) might calculate \(18 \times 15 = 270\) and forget to divide by 2 to account for double-counting, leading them to the incorrect answer of 270.

Errors while selecting the answer

No likely faltering points - once students have correctly calculated 135 handshakes, the answer selection is straightforward as it directly matches choice B.