Question

Of the 5 distinguishable wires that lead into an apartment, 2 are for cable television service and 3 are for telephone service. Using these wires, how many distinct combinations of 3 wires are there such that at least 1 of the wires is for cable television?

Short answer

There are 9 distinct combinations of 3 wires such that at least 1 wire is for cable television.

Step-by-step solution

- Total Combinations of 3 Wires

First, determine the total number of ways to select 3 wires out of 5 without considering the type of wire. Use the combination formula \( \binom{n}{k} \)) where \( n\) is the total number of wires and \( k\) is the number of wires to choose: \[ \binom{5}{3} = \frac{5!}{3!(5-3)!} = 10 \]

- Combinations of 3 Telephone Wires

Next, determine how many ways you can select 3 telephone wires out of the 3 available. This ensures all selected wires are for telephone: \[ \binom{3}{3} = \frac{3!}{3!(3-3)!} = 1 \]

- Subtract Unfavorable Combination

To find the number of combinations with at least 1 cable wire, subtract the all-telephone combination from the total combinations calculated in Step 1: \[ \text{Total combinations} - \text{All-telephone combinations} = 10 - 1 = 9 \]

Key concepts

Combinations

The concept of combinations is used to determine how many ways you can choose items from a group without considering the order of selection. In combinatorics, this is crucial for solving problems like the one given, where the arrangement doesn't matter. The formula for combinations is given by \( \binom{n}{k} \), which means choosing \( k \) items from \( n \) items and is calculated as: \[ \binom{n}{k} = \frac{n!}{k!(n-k)!}\ \] In our problem, we needed to find how many ways we can select 3 wires out of 5 distinguishable wires: \[ \binom{5}{3} = \frac{5!}{3!(5-3)!} = 10 \ \] This gives us the total combinations of 3 wires from the 5. Always remember that combinations disregard the order in which objects are chosen. Whether you pick wire1, wire2, wire3 or wire3, wire1, wire2, it's counted as the same combination.

Probability

Probability deals with quantifying the chance or likelihood of an event occurring. In this problem, the interest was to find the number of suitable combinations where at least one of the wires was for cable television out of the selected 3 wires. First, we calculated the total possible combinations, which were found to be 10. The next step involved subtracting the unfavorable situations. The only unfavorable situation here was selecting all telephone wires: \[ \binom{3}{3} = \frac{3!}{3!(3-3)!} = 1 \] Therefore, there is only one way to pick 3 wires, all being for the telephone service. To find the combinations where at least one wire is for cable television, we had to subtract this from the total combinations: \[ 10 - 1 = 9 \] Hence, there are 9 favorable combinations where at least one wire is for cable television.

GMAT Problem Solving

The GMAT often features combinatorics and probability questions, requiring a methodical approach to problem-solving. Here's how you can tackle similar problems:

• Carefully read and understand what the problem is asking.
• Identify and separate different cases or groups within the problem (e.g., cable vs. telephone wires).
• Use appropriate mathematical tools (like combinations) to calculate different cases.
• Consider complementary counting (subtracting unfavorable cases from the total) to simplify your solution.
• Double-check your steps to make sure no unfavorable scenarios are missed.

By following these steps, not only can you sail through combinatorics questions but also build a strong foundation for more complex GMAT problems. Practice regularly to familiarize yourself with different problem types and improve your speed and accuracy.