OpenStax-CNX module: m37202
1
Unordered Sampling Without
Replacement
∗
Avinash Hathiramani
This work is produced by OpenStax-CNX and licensed under the
†
Creative Commons Attribution License 3.0
Abstract
This article introduces the method of calculating the number of unordered arrangements of objects
without replacement.
It is a brief article which attempts to explain this concept from an intuitive
perspective making it easy to understand and derive the formulae.
1 Unordered Sampling Without Replacement
2 The Basic Case
Unordered sampling without replacement1 can be thought of as determining how many ways there are to
divide n objects into r groups. Here we are choosing the number of dividers of the object (not the number
of objects per se). For example, if we have seven objects which we want to divide, we choose how many
dividers to put between the objects:
O | O O | O O O | O
With 3 dividers we can put the objects into 4 groups, i.e. we only specify r − 1 dividers. There are n − 1
spaces between the objects, so n − 1 ways in which to choose these r − 1 dividers.
Note this is specically if each group must contain at least one object.
n−1
So this gives the result that there are Cr−1
distinct positive (note not non-negative) integer-valued vectors
(n1 , n2 , · · · , nr ) satisfying the condition n1 + n2 + · · · + nr = n. (Remember this is for nk ≥ 1, ∀k : k ∈ [1, r]).
Notice the result and bear in mind the conditions
n−1
Cr−1
,
for n1 + n2 + · · · + nr = n,
where nk ≥ 1.
So in general the sum of the individual groups gives the value to be used in the top position of the
combination formula.
3 The Generalised Case
Now we turn to the situation where we have nk ≥ j .
If we can get the sum of a series that satised the condition that each term in that series is ≥ 1 then we
will be able to substitute the sum of this series for n in the combination formula given above.
Note that although we have changed the condition from nk ≥ 1 to nk ≥ j , we must still have the
condition that n1 + n2 + · · · + nr = n, otherwise we could end up with a dierent number of objects than
we started with after dividing them into the groups.
∗ Version
1.5: Mar 28, 2011 4:34 pm -0500
† http://creativecommons.org/licenses/by/3.0/
1 Also
known as the `balls in boxes' or `balls in urn' problem.
http://cnx.org/content/m37202/1.5/
OpenStax-CNX module: m37202
2
So what term is ≥ 1 when nk ≥ j ?
What we have to start with is nk ≥ 1. So if nk ≥ 1, this means that nk −j ≥ 0, and therefore nk −j+1 ≥ 1.
All that we need to do now is to sum this series and substitute the result of this sum for n in the
combination formula given above.
The sum of this series is
(n1 − j + 1) + (n2 − j + 1) + · · · + (nr − j + 1)
(1)
Rearranging so that all of the terms in n are together, and summing the common (−j + 1) term r-times,
we get
(n1 + n2 + · · · + nr ) + (−j + 1) r
(2)
as the sum of our series.
Now, recall that we still have the condition that n1 + n2 + · · · + nr = n. Using this fact, and rewritting
(−j + 1) as (1 − j), we get
n + (1 − j) r
(3)
as the sum of our series.
We can substitute this sum into the combination formula from "The Basic Case " (Section 2: The Basic
Case ) (in place of the n term in this formula). So that
n+(1−j)r−1
n−1
Cr−1
7→ Cr−1
(4)
Giving the nal result of this paper, that:
n+(1−j)r−1
There are Cr−1
number of ways to group n objects into r groups when nk ≥ j 2 and n1 + n2 +
3
· · · + nr = n .
4 References
Ross, Sheldon, A First Course in Probability, Eigth Edition, Pearson Education, 2008.
Special Thanks to Dr. Stephen Connor from the University of York, on whose notes this article is based.
2 I.e.
3 I.e.
We must have at least
j
objects in each group.
We must allocate exactly the number of objects that we started with in the beginning.
http://cnx.org/content/m37202/1.5/