An order of award presentations has been devised for seven people: Jeff, Karen, Lyle, Maria, Norm, Olivia, and Paul.

In how many ways can the first award be presented to Karen and the last to Lyle?

Question

An order of award presentations has been devised for seven people: Jeff, Karen, Lyle, Maria, Norm, Olivia, and Paul.

In how many ways can the first award be presented to Karen and the last to Lyle?

BlackTom AI · Accepted Answer

The problem sets a fixed position for the first and last awards and asks how many ways the remaining positions can be filled.
First, Karen must receive the first award and Lyle must receive the last award. That fixes two of the seven people in place.
This leaves five people—Jeff, Maria, Norm, Olivia, and Paul—to fill the five interior positions (positions 2 through 6) in any order.
The number of possible arrangements for these five people is 5! (five factorial), which equals 120.
Now evaluate the answer choices: 
- 840: This would correspond to a larger count, such as 7! divided by some constraint, but with two fixed ends the count is not that high.
- 24: This is 4!, which would imply fixing four interior positions instead of five, so it’s too small.
- 120: This matches 5!, the correct count for the interior permutations given the two fixed endpoints.
- 360: This is 3! times 5 or similar, but it does not align with fixing two endpoints and permuting five interiors.
Thus, the only option consistent with the described constraints is 120.

An order of award presentations has been devised for seven people: Jeff, Karen, Lyle, Maria, Norm, Olivia, and Paul. In how many ways can the first award be presented to Karen and the last to Lyle?单项选择题

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