Let us also number the audits from 1 to 9.

This is a combinations problem because the order does not matter.

The formula for combinations is

C(n,r) = n! / (r! (n-r)!)

where n is the total possibilites and the r is the things chosen from n possibilities

Accountant A

C(9,4) = 9! / (4! (9-4)! ) =

9*8*7*6*5*4*3*2*1

————————– = 126

4*3*2*1*5*4*3*2*1

Accountant B

C(9,3) = 9! / ( 3! (9-3)! ) =

9*8*7*6*5*4*3*2*1

————————– = 84

3*2*1*6*5*4*3*2*1

Accountant C

C(9,2) = 9! / (2! (9-2)!) =

9*8*7*6*5*4*3*2*1

————————– = 36

2*1*7*6*5*4*3*2*1

From the above, it can be seen that Accountant A can be assigned 4 audits in a possible of 126 ways. Accountant B can be assigned 3 audits in a possible of 84 ways and finally, Accountant C can be assigned 2 audits in a possible of 36 ways.

A does 2, B does 3, C does 4

A does 2, B does 4, C does 3;

A does 3, B does 2, C does 4

A does 3, B does 4, C does 2

A does 4, B does 2, C does 3

A does 4, B does 3, C does 2.