Step-by-step Answer:
We will discuss first the second problem, since the first is the same nature, but with bigger numbers.
2. If I have 15 doctors and there are four offices, how many ways can I arrange them so that only only one doctor can hold one office? <—permutation
If the offices are numbered, the first office has a choice of 15 candidates, the second has 14, the third has 13, and the fourth has 12 for a total of
N=15*14*13*12 = 32760 arrangements.
This number is also represented easily as
N=C(15,1)*C(14,1)*C(13,1)*C(12,1) = 32760
where C(n,r) = n!/((n-r)!r!)
However, if the offices are not numbered, or if order of choice of doctors is not important, then we have to divided N by the number of ways the four offices than can be arranged, namely 4*3*2*1=24, and the number of arrangements is reduced by
n = N/24 = 1365 = C(15,4), i.e. number of ways we can choose a group of four doctors (order does not count) from 15.
1. Similarly for 40 people grouped into four groups of 10.
If the groups are numbered, then the number of arrangements is
N=C(40,10)*C(30,10)*C(20,10)*C(10,10) = 4705360871073570227520
If the groups are not numbered, the number of arrangements is therefore
n=N/4! = 4705360871073570227520 / 24 = 196056702961398759480
