###### Univariate vs. Multivariate Distributi ...

Univariate and multivariate normal distributions are very robust and useful in most statistical... **Read More**

Counting problems involve determining the exact number of ways two or more operations or events can be performed simultaneously. For instance, we might be interested in the number of ways to choose 7 chartered analysts comprising 3 women and 4 men from a group of 50 analysts. Counting encompasses the following fundamental principles:

“\(n\) factorial” (\(n!\)) is used to represent the product of the first \(n\)* ***natural numbers**. Generally:

$$ n! = n \times (n – 1) \times (n – 2) \times (n – 3) \times … \times 2 \times 1 $$

For example,

\(1! = 1 \times 1\).

\(2! = 2 \times 1\).

\(3! = 3 \times 2 \times 1\).

\(4! = 4 \times 3 \times 2 \times 1\).

* Note to candidates*: 0! is just 1, not zero. You should also remember that we can only find \(n!\) if \(n\) is a

The labeling principle is used to assign \(k\) labels or groups to a total of \(n\) items, where each label contains \(n_i\) items such that \(n_1+n_2+n_3 + … +n_k = n\). In other words, your wish is to have \(n\) items categorized into \(k\) groups, where the number of items in each group is **pre-determined**. To get the total number of ways that the labels or groups can be assigned, you use the formula:

$$ \text{Number of labels} = \cfrac { n! }{ { n }_{ 1 }\times{ n }_{ 2 }\times{ n }_{ 3 }\times … \times { n }_{ k } }

$$

Assume that you have a portfolio of investments consisting of 10 stocks. Suppose your wish is to assign 3 different labels such that label 1 has 5 “high return” stocks, label 2 has 3 “medium return” stocks, and the last label has 2 “low return” stocks:

**Solution**

\(n = 10\).

There are 3 labels, where \(n_1 = 5\), \(n_2 = 3\), and \(n_3 = 2\).

The number of different ways that you can assign the 3 labels =\(\cfrac {10!}{(5! \times 3! \times 2!)} = 2520 \text{ ways}.\)

A combination is a selection of some given items where the order **does not** matter. The number of combinations (possible ways) of \(n\) items taken \(r\) at a time is:

$$ \text{Number of combinations}, \text{ nCr}=\cfrac { n! }{ \left( n-r \right) !r! } $$

How many ways can we choose 3 stocks from a portfolio of 10 stocks?

**Solution**

We are interested in the number of combinations of 10 items taken 3 at a time. Therefore,

\(n = 10\).

\(r = 3\).

Number of possible combinations =\(\cfrac {10!}{(7! \times 3!)} = 120 \)

Unlike a combination, a permutation involves determining the number of possible ways to choose \(r\) items from \(n\) items. In a permutation, **the** **order is paramount**. Simply put**, **the order of the \(r\) items chosen matters, i.e., which one comes first? Which one should come last?

$$ \text{Number of permutations}, \text{ nPr}=\cfrac { n! }{ \left( n-r \right) ! } $$

Let us refer to example 2. Assume that the three chosen stocks are to be sold in an arrangement in which order of sale is important.

**Solution**

This means that once we have chosen 3 stocks, we must also determine the order in which to sell them. Therefore, the number of possible permutations \(= 10!/7! = 720\).

* Note to candidates*: If you compare the combination formula to the permutation formula, the only difference is the \(r!\) in the denominator of the former. This means that in any situation, there are always \(r!\) more ways to choose items

- If you are asked to assign \(n\) items to (\n\) slots, use the factorial formula.
- In case you are asked to assign \(k\) unique labels or categories to \(n\) items, use the labeling formula.
- When asked to come up with a number of ways to choose \(r\) items from \(n\) items when the order is not important, use the combination formula. If the order is important, use the permutation formula.

QuestionA company has 10 male and 18 female employees. The company chooses 6 employees at random for deployment to another recently opened branch.Calculate the probability that the chosen employees consist of 3 males and 3 females.

- 0.18.
- 0.25.
- 0.26.

SolutionThe total number of ways of choosing 6 employees is given by:

$$ \binom{28}{6}=376,740 $$Now, the number of ways of choosing 3 males and 3 females in their respective groups is given by:

$$ \binom{10}{3}\cdot \binom{18}{3}=97,920 $$. Thus, the probability of choosing 3 males and 3 females is given by:

$$ \frac{97,920}{376,740}=0.26 $$