The number of arrangements of 10 different things taken 4 at a time is given by the formula:

nPr = n! / (n-r)!

where n is the total number of things, and r is the number of things taken at a time.

n = 10 and r = 4, so:

nPr = 10! / (10-4)! = 10! / 6! = 10 × 9 × 8 × 7 = 5,040

Therefore, there are 5,040 total arrangements of 10 different things taken 4 at a time.



Since one particular thing always occurs in the arrangement, we can fix its position and arrange the remaining 3 things from the remaining 9 things.

This can be done in nPr ways, where n is the total number of remaining things and r is the number of things taken at a time.

n = 9 and r = 3, so:

nPr = 9! / (9-3)! = 9! / 6! = 9 × 8 × 7 = 504

Therefore, there are 504 arrangements including one particular thing.



Since one particular thing is not included in the arrangement, we can choose 4 things from the remaining 9 things.

This can be done in nCr ways, where n is the total number of things and r is the number of things taken at a time.

n = 9 and r = 4, so:

nCr = 9! / (4! × (9-4)!) = 9! / (4! × 5!) = 9 × 8 × 7 × 6 / (4 × 3 × 2 × 1) = 126

Therefore, there are 126 arrangements without one particular thing.



The number of arrangements with one particular thing always occurring is the product of the number of arrangements including one particular thing and the number of arrangements without one particular thing.

Therefore, the number of arrangements with one particular thing always occurring is:

504 × 126 = 63,504

Therefore, there are 63,504 arrangements of 10 different things taken 4 at a time in which one particular thing always occurs.