If a club has 12 members, what is the ratio of the number of different possible 6-member committees that can...

1. Translate the problem requirements: We need to find the ratio of \(\mathrm{C(12,6)}\) to \(\mathrm{C(12,5)}\), where \(\mathrm{C(n,r)}\) represents the number of ways to choose r items from n items without regard to order.
2. Apply the complementary combinations property: Recognize that choosing 6 members from 12 is the same as choosing 6 members to leave out, which equals choosing 6 from 12, so \(\mathrm{C(12,6)} = \mathrm{C(12,6)}\).
3. Use the relationship between adjacent combinations: Apply the identity that \(\mathrm{C(n,r)} = \mathrm{C(n,n-r)}\) and the recursive relationship to find that \(\mathrm{C(12,6)}\) relates to \(\mathrm{C(12,5)}\) through the remaining members.
4. Calculate the ratio using the combination relationship: Use \(\frac{\mathrm{C(12,6)}}{\mathrm{C(12,5)}} = \frac{12-5}{6} = \frac{7}{6}\) by applying the property that \(\frac{\mathrm{C(n,r)}}{\mathrm{C(n,r-1)}} = \frac{n-r+1}{r}\).

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're actually looking for. We have a club with 12 members, and we want to compare two things:

• How many different ways can we form a committee of 6 members?
• How many different ways can we form a committee of 5 members?

Then we need to find the ratio between these two numbers.

In mathematical terms, we're looking for ways to choose people from a group where the order doesn't matter - this is called a combination. We write this as \(\mathrm{C(12,6)}\) for choosing 6 people from 12, and \(\mathrm{C(12,5)}\) for choosing 5 people from 12.

Our goal is to find the ratio \(\mathrm{C(12,6)} : \mathrm{C(12,5)}\).

Process Skill: TRANSLATE - Converting the word problem into clear mathematical understanding

2. Apply the complementary combinations property

Here's a key insight that makes this problem much easier: when you choose 6 members to be ON a committee from 12 people, you're automatically choosing 6 members to be OFF the committee.

Think about it this way: if I pick 6 people to include, then the remaining 6 people are automatically excluded. So choosing 6 people to include is exactly the same as choosing 6 people to exclude.

This means: \(\mathrm{C(12,6)} = \mathrm{C(12,6)}\)

But more usefully, we can also say: choosing 6 people from 12 is the same as choosing which 6 people to leave out, which is the same as choosing 6 from 12.

Actually, the most helpful relationship is: \(\mathrm{C(12,6)} = \mathrm{C(12,12-6)} = \mathrm{C(12,6)}\)

This complementary property will help us see the relationship with \(\mathrm{C(12,5)}\).

3. Use the relationship between adjacent combinations

Now let's think about the relationship between \(\mathrm{C(12,6)}\) and \(\mathrm{C(12,5)}\) in an intuitive way.

When we form a 5-member committee, we're leaving 7 people out. When we form a 6-member committee, we're leaving 6 people out.

Here's the key insight: every 5-member committee can be connected to several 6-member committees. Specifically, for each 5-member committee, we can form 6-member committees by adding any one of the 7 remaining people.

So each 5-member committee leads to exactly 7 different 6-member committees.

But wait - we need to be careful about double-counting. When we go from 6-member committees back to 5-member committees, each 6-member committee can be reduced to a 5-member committee in exactly 6 different ways (by removing any one of the 6 members).

This gives us the relationship we need to find the ratio.

4. Calculate the ratio using the combination relationship

From our reasoning above, we can establish the relationship between these combinations.

There's a direct formula for the ratio of adjacent combinations:
\(\frac{\mathrm{C(n,r)}}{\mathrm{C(n,r-1)}} = \frac{n-r+1}{r}\)

In our case, we want \(\frac{\mathrm{C(12,6)}}{\mathrm{C(12,5)}}\).
Using the formula with n=12, r=6:

\(\frac{\mathrm{C(12,6)}}{\mathrm{C(12,5)}} = \frac{12-6+1}{6} = \frac{7}{6}\)

Let's verify this makes sense with our intuitive reasoning:

• Every 5-member committee can become 7 different 6-member committees (by adding any of the 7 excluded members)
• Every 6-member committee comes from 6 different 5-member committees (by removing any of the 6 included members)
• So there are \(\frac{7}{6}\) times as many 6-member committees as 5-member committees

Therefore, the ratio is 7 to 6.

4. Final Answer

The ratio of the number of different possible 6-member committees to the number of different possible 5-member committees is 7 to 6.

Looking at our answer choices, this matches option E: 7 to 6.

Answer: E

Common Faltering Points

Errors while devising the approach

1. Confusing combinations with permutations

Students often struggle to recognize that this is a combinations problem where order doesn't matter. They might think that selecting members A, B, C, D, E, F is different from selecting F, E, D, C, B, A for a committee, leading them to use permutation formulas instead of combination formulas.

2. Misunderstanding what ratio is being asked

The question asks for the ratio of 6-member committees to 5-member committees, but students might accidentally calculate the ratio in reverse (5-member to 6-member) or get confused about which number should be the numerator and which should be the denominator.

3. Attempting to calculate actual values instead of using relationships

Students might think they need to calculate the exact values of \(\mathrm{C(12,6)}\) and \(\mathrm{C(12,5)}\) using the full combination formula, not recognizing that there's a more elegant relationship between adjacent combinations that can give the ratio directly.

Errors while executing the approach

1. Arithmetic errors in the combination relationship formula

When using the formula \(\frac{\mathrm{C(n,r)}}{\mathrm{C(n,r-1)}} = \frac{n-r+1}{r}\), students often make substitution errors. For example, they might incorrectly substitute values as \(\frac{12-5+1}{5}\) instead of \(\frac{12-6+1}{6}\), or make basic arithmetic mistakes like calculating 12-6+1 as 6 instead of 7.

2. Misapplying the complementary combination property

Students might correctly identify that \(\mathrm{C(12,6)} = \mathrm{C(12,6)}\) due to the complementary property, but then incorrectly try to use \(\mathrm{C(12,7)}\) instead of \(\mathrm{C(12,5)}\) in their ratio, not understanding which complementary relationship actually helps solve the problem.

Errors while selecting the answer

1. Expressing the ratio in wrong order

After correctly calculating that the relationship gives \(\frac{7}{6}\), students might select '6 to 7' instead of '7 to 6' because they flip the ratio when looking at the answer choices, especially if they lose track of which committee type was supposed to be first in the ratio.