On the board of a club there are 8 first class petty officers and 6 second class petty officers. In how many different ways can a president, vice president, secretary, and treasurer be chosen if the president and secretary must both be first class petty officers and the vice president and treasurer must both be second class petty officers?

Introduction / Context:
This question tests your ability to count arrangements when positions have specific eligibility constraints. We are choosing four different posts, each from a restricted subset of people. Because the order of posts clearly matters (president is not the same as secretary), we must use permutations inside restricted groups. These kinds of problems are common in exam questions that mix combinatorics and logical thinking.

Given Data / Assumptions:

• First class petty officers: 8 people.
• Second class petty officers: 6 people.
• Positions: president, vice president, secretary, treasurer.
• President and secretary must be first class petty officers.
• Vice president and treasurer must be second class petty officers.
• No person can hold more than one post simultaneously.

Concept / Approach:
We break the problem into two independent parts:

• Assign president and secretary from the 8 first class petty officers.
• Assign vice president and treasurer from the 6 second class petty officers.

Within each group, the positions are distinct, so we use permutations. Finally, we multiply the counts from both groups because choosing first class posts and second class posts are independent tasks.

Step-by-Step Solution:
Step 1: Number of ways to choose a president from 8 first class officers = 8. Step 2: After choosing the president, 7 first class officers remain for the secretary post. Step 3: Ways to assign president and secretary from first class officers = 8 * 7 = 56. Step 4: Now consider second class officers for vice president and treasurer. Step 5: Number of ways to choose a vice president from 6 second class officers = 6. Step 6: After that, 5 officers remain for treasurer, so ways = 6 * 5 = 30. Step 7: Total ways = ways to assign first class posts * ways to assign second class posts = 56 * 30 = 1680.

Verification / Alternative check:
The same calculation can be expressed as permutations: for first class posts, 8P2 = 8 * 7 = 56, and for second class posts, 6P2 = 6 * 5 = 30. The total number of assignments is 8P2 * 6P2 = 56 * 30 = 1680. Both approaches match perfectly, confirming the result is consistent and correct.

Why Other Options Are Wrong:
1500: Does not match the exact product 56 * 30 and indicates a miscalculation. 1860: Another incorrect multiplication, possibly from mixing 8C2 or 6C2 with permutations. 1640: Close but still incorrect; it does not arise from any valid combination of permutations here. Only 1680 correctly reflects the separate permutations for first class and second class roles.

Common Pitfalls:
Some learners treat the four posts as identical, incorrectly using combinations instead of permutations. Others ignore the class restrictions and choose all four officers from the combined pool without distinguishing first class and second class categories. It is also easy to forget that once a person is selected for one role, they cannot be chosen again for another, leading to overcounting. Always carefully interpret role constraints and apply permutations where order (which post is held) clearly matters.

Final Answer:
The number of different ways to choose the four officers under the given conditions is 1680.