the number of ways in which 18 students in Mr. Garr grade 3 class can line up for a photo if the the Armstrong triplets cannot all be together is n x16!
what is n?
To find the value of "n," we need to determine the number of ways in which the Armstrong triplets can be together in line for the photo. Once we calculate that, we can subtract it from the total number of ways to rearrange all 18 students in Mr. Garr's grade 3 class.
Let's first calculate the number of ways the Armstrong triplets can be together. Since the triplets need to be treated as a single unit, we can consider them as one element. So, we have 16 elements to arrange (15 individual students + 1 group of triplets).
The number of ways to arrange these 16 elements can be given by 16!.
Next, we need to account for the different ways the triplets can be arranged within the group. Since the triplets can be arranged amongst themselves in 3! (3 factorial) ways, we need to multiply 16! by 3!.
Therefore, the number of ways the Armstrong triplets can be together is 16! * 3!.
Finally, to find "n," we need to subtract the number of arrangements with the triplets from the total number of ways to arrange all 18 students.
The total number of ways to arrange the 18 students can be given by 18!.
Hence, n = 18! - (16! * 3!).