When John arrives in New York, he has 8 different tourist stops that he could visit, but he has time to visit only 6 of them. If the order in which he visits the 6 stops matters (it is a schedule), then in how many different ways can he choose and arrange the 6 stops?

Introduction / Context:
This problem asks us to count how many different itineraries John can follow if he chooses and orders a subset of tourist stops. Because both the choice of which locations he visits and the order in which he visits them matter, we are dealing with permutations of a subset of the available stops. This kind of question appears frequently in arrangement and scheduling contexts in aptitude tests.

Given Data / Assumptions:

• Total distinct tourist stops available = 8.
• John will visit exactly 6 stops.
• The order of visiting the stops is important.
• No stop is visited more than once.

Concept / Approach:
We need the number of permutations of 8 distinct items taken 6 at a time. This is denoted by P(8, 6) and equals 8! divided by (8 - 6)!, that is 8! / 2!. Conceptually, John first chooses 6 stops from the 8, and then arranges these 6 in all possible orders, so the result captures both selection and ordering.

Step-by-Step Solution:
Step 1: Use the permutation formula for selecting and ordering 6 stops out of 8. P(8, 6) = 8! / (8 - 6)! = 8! / 2!. Step 2: Compute 8! = 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 40320. Step 3: Compute 2! = 2. Step 4: Divide to obtain the number of permutations. P(8, 6) = 40320 / 2 = 20160.

Verification / Alternative check:
Alternatively, you can reason position by position. For the first stop, John can choose any of 8 places. For the second stop, 7 choices remain, then 6 for the third, and so on down to 3 choices for the sixth stop. The product is 8 * 7 * 6 * 5 * 4 * 3, which equals 20160. This matches the permutation approach and confirms the final answer.

Why Other Options Are Wrong:
Values such as 20610, 21000 or 24000 are close to 20160 but not equal to the exact product 8 * 7 * 6 * 5 * 4 * 3. They indicate arithmetic mistakes or using extra or fewer factors. The value 13440 would arise if only five positions were considered or if some choices were miscounted. None of these incorrect options equals P(8, 6).

Common Pitfalls:
One frequent mistake is to treat the problem as a combination and use C(8, 6), which counts only which stops are visited, not their order. Another error is to truncate the product too early or incorrectly include extra multiplication factors. By clearly recognising that the question involves both selection and ordering of stops, and by writing out the full product, we avoid these pitfalls.

Final Answer:
John can choose and arrange his 6 stop schedule in 20160 different ways.