“Multinomial Coefficient”: A Generalization of the Binomial Coefficient
Example:
A team plays 16 games in a season. At the end of the season, the team has 8 wins, 3 ties
and 5 losses. How many different seasons are possible? That is, how many different ways
are there to order the 8 wins, 3 ties and 5 losses, when all outcomes of a given type (win, tie
or loss) are considered identical (i.e. we don’t distinguish between two wins).
Soln.
Let’s line up 16 spots for the wins, losses and ties:
Let W , T and L, denote a win, a tie and a loss, respectively. The number of ways to order
the wins, ties and losses is the number of sequences of 8 wins, 5 ties and 3 losses. How many
such sequences are there?
To answer this question using the basic counting tools, let’s think of the complex task of
constructing a “word” with W 0 s, T 0 s, and L0 s as a series of three simple sub-actions and
then apply the multiplication principle. One choice for the first sub-action is to choose 8
spots from the available 16 for the wins. A choice for the second sub-action, then, is to pick
3 positions from the remaining 8 for the ties. It follows that the third sub-action is to put
the 5 losses into the 5 remaining spots.
How many ways are there to accomplish each of the three sub-actions? The number
of ways to do sub-action 1 is the number of ways to choose an unordered sample of eight
games from the 16 (distinguishable)
games without replacement, so the number of ways to
do sub-action 1 is C(16, 8) = 16
. Now, there are 8 available positions for the 3 ties. By
8
a completely analogous argument, the number of ways to do sub-action 2 is C(8, 5) = 85 .
Finally, after completing the first two sub-actions, only 5 games remain for the 5 losses.
Therefore, there is only 1 way to do sub-action 3. (We could also argue, in analogy with
the first two sub-actions, that there is C(5, 5) = 1 way to do sub-action 3, and arrive at the
same conclusion).
By the multiplication principle, there are
16 8
16!
=
8
5
8!8!
possible seasons.
The decision to pick the games for the wins first, the ties second and the losses third is
1
arbitrary. You can check that you arrive at the same answer for the number of possible
seasons if you order the sub-actions in a different way, such as picking 5 spots for the losses
first followed by 3 spots for the ties, or picking 3 spots for the ties first followed by 8 spots
for the wins, etc.
The number of possible seasons is an example of a “multinomial coefficient.”
Definition: Multinomial Coefficient - The number of distinguishable ways to
order n objects, where
Pk n1 are of type 1, n2 are of type 2, . . ., and nk are of
type k, where n = i=1 ni .
Theorem: The total number of distinguishable ways to order n objects, where
n1 are of type 1, n2 are of type 2, . . ., and nk are of type k is
n!
,
n1 ! · · · nk !
where n =
Pk
i=1 ni .
Note: The binomial coefficient is a special case of the multinomial coefficient when there are exactly two types of objects. In this case, there are
n1 objects of type 1 and n2 = n − n1 objects of type 2. The multinomial
coefficient, defined above, is the same as the binomial coefficient:
n!
n!
n
=
=
.
n1 !n2 ! n1 !(n − n1 )!
n1
Example...continued...If all possible outcomes (sequences of wins, ties, and
losses) are equally likely (unrealistic, but we can pretend...), what is the
probability of three consecutive ties?
# ways to have three consecutive ties
# possible seasons
13 5
14 8 5
= .025.
=
16 8
P (exactly 3 ties) =
8
5
2
An Application of Unordered Samples Without Replacement
(Reference: Larsen and Marx, pg. 128-130)
The Situation: A box has N objects; r objects are of type 1, and N − r objects are of
type 2. We randomly select n objects from the box without replacement.
The Question: What is the probability that we select k objects of type 1?
A Solution:
First, what is Ω? An outcome consists of a combination of n objects chosen from N ,
so Ω = { unordered samples of size n taken from N without replacement}.
Let A be the event that we select k objects of type 1. Since all outcomes in Ω are
equally likely, P (A) = |A|
.
|Ω|
What is |Ω|? The number of outcomes in Ω is the total number of combinations of n
distinguishable objects chosen from N distinguishable objects; |Ω| = Nn .
What is |A|? If k > r or k > n, then |A| = 0. Suppose k ≤ min{r, n}. To figure
out |A|, break the action of selecting k objects of type 1 into two sub-actions:
• Sub-action 1: Select k objects from
the r objects that are of type 1. The number of
ways to perform sub-action 1 is kr
• Sub-action 2: Select n − k objects from the N − r objects that are of type 2. The
−r
number of ways to perform sub-action 2 is Nn−k
−r
By the multiplication principle, |A| = kr Nn−k
. Then, for k = 0, . . . , min{r, n},
r N −r
|A|
k
n−k
.
P (A) =
=
N
|Ω|
r
3
We can summarize the above with the following general result.
Result: Suppose that in a box with N objects, r of the objects are of type 1
and N − r of the objects are of type 2. We randomly select n ≤ N objects
from the box without replacement. Then,
P( select k objects of type 1) =
 r N −r
(k)( n−k )


 (Nr )
if k = 1, . . . , min{r, n}



0
otherwise.
4
Example: Acceptance Sampling (Larsen and Marx, pg. 132-133)
A computer manufacturer orders a supply of parts. The manufacturer would like to have confidence
that the parts satisfy certain previously specified requirements (i.e. no defects). In principle, the manufacturer could inspect every single part in the lot. In practice, however, practical constraints such as limited
time and money prohibit such a thorough investigation. A more efficient solution is to select a random
sample of the parts, inspect only the sampled parts and accept the shipment only if the quality of the sample
is sufficiently high.
Suppose a shipment contains N = 100 parts. The manufacturer only accepts a
shipment if the sample contains no defective parts.
1. Suppose a sample of size n = 2 is selected for inspection. What is the probability that a shipment with
10% defects is accepted?
2. If the manufacturer wants to have less than a 50% chance of accepting a 10% defective shipment, how
big of a sample should the manufacturer select?
Solution
The shipment consists of two types of objects: defective parts and parts that are not defective. Let r
be the number of defective items in the shipment. The manufacturer selects a sample of size n of the parts.
Let the event A be the event that the manufacturer selects k defective items in a sample of size n. Using
the above “result,”
r 100−r
P (A) =
k
n−k
100
n
.
P (A) is a function of r, n, and k.
For the first question, the sample size is n = 2, and the number of defectives is r = 10. A shipment is accepted only if there are no defective parts in the sample. To compute the probability in question,
we compute P (A), as shown above, with k = 0. This gives
10 90
P (accept sample ) =
0
2
100
2
= .809.
The plot in Figure 1 shows the probability of accepting a shipment as a function of the number of defective
items when the sample size is 2. The star in the graph indicates that the probability of accepting a shipment
with 10 defective items is approximately .809.
For question 2, the manufacturer wants to know how big the sample size needs to be for the probability of accepting a shipment with 10 defective items to be less than .5. The plot in Figure 1 shows the
probabilities of accepting a shipment when when there are 10 defective items for sample sizes between 1 and
30. That is, the number of defectives, r, is fixed at 10, and the sample size ranges from n = 1 to n = 30. The
horizontal line in the plot is at a probability of .5. Since the dots in the plot, first fall below the horizontal
line at the sample size of n = 7, the manufacturer needs to select a sample of at least size seven to have less
than a 50% chance of accepting a 10% defective shipment.
5
Figure 1: Plot for Question 1.
6
Figure 2: Plot for Question 2.
7