it would be a combination because order is not necessary.
this particular problem has more possible combinations than my calculator can handle.
i went to excel.
excel was able to handle it.
it told me that c(1111,99) = 3.9879E+143
that's the number of combinations.
it told me that p(1111,99) = 3.7218E+299
that's the number of permutations.
the formula for number of combinations is c(n,x) = n! / (x! * (n-x)!)
the formula for number of permutations is p(n,x) = n! / (n-x)!
the difference is the division by x! which is the number of arrangements in the smaller set of 99.
to let you understand the difference, i chose much smaller numbers.
a locker contains 3 books.
you select 2 books at random.
how many different sets of books can you select?
c(3,2) = 3! / (2! * 1!) = 3
p(3,2) = 3! / 1! = 6
let the books be labeled a,b,c
the number of possible combinations is:
a,b
a,c
b,c
the number of possible permutations is:
a,b
b,a
a,c
c,a
b,c
c,b
with c(3,2), the order of the elements inside the smaller set doesn't matter, so a,b and b,a are both considered to be part of the same set.
so are a,c and c,a.
so are b,c and c,b
with p(3,2), the order of the elements inside the smaller set does matter, so a,b is considered a different set than b,a.
note that, if you have a set of 2 elements and you want to know the possible arrangements that can be made of that set, you would multiply that set by 2!.
by taking the permutation formula of n! / (n-x)!) and you want to remove the possible arrangements in the resulting permutation, you just divide by x!.
that gets you the combination formula of n! / (x! * (n-x)!)
from the excel results, you can see that the number of permutations is much greater than the number of combinations.
when you divide the number of permutations by 99!, you get the number of combinations, as shown below.
3.9879E+143 ***** equals c(1111,99) equals number of combinations.
3.7218E+299 equals p(1111,99) equals number of permutations.
9.3326E+155 equals 99! equals 99 factorial.
3.9879E+143 ***** equals p(1111,99) / 99! equals number of permutations
divided by 99!.
***** p(1111,99) / 99! is the same as c(1111,99).
a breakdown of the formulas might also show you the difference.
p(3,2) = 3! / 1! = 3 * 2 * 1 = 6
c(3,2) = 3! / (1! * 2!) = (3 * 2 * 1 / (1 * 2) = 3.
the permutation formula gave you 3 * 2!.
the 1! is the number of possible arrangements in a set of 2 elements.
the combination formula gave you 3 without multiplying by 2!, so you only get the elements without any ordering of them.
you only wanted to know the number of possible sets of 99.
no order within the set of 99 was implied, so it's a combination.
if you had wanted to know the number of possble sets of 99 and the number of possible arrangements of those books within each set of 99, then it would be a permutation.