Probability Permutation Examples

# Probability Permutation Examples

Consider a set of different items. The items may be anything, say numbers, colors, names, designations etc. We then have an inquisitiveness that in how many ways the set of items can be arranged. We use a fundamental approach, called as principle of counting and find the number of arrangements. Again, there are two possible types of arrangements. One is by considering order and the other is not considering the order at all. To understand this, let us consider a simple example.

Let given set consists of three different letters. What are the different ways of arranging these letters? When the order is considered as important, the possible number of arrangements is $6$, namely $ABC, ACB, BCA, BAC, CAB, CBA$. But if one does not worry about the order, then, to him all these six arrangements merge into a single one. In practice, we say that the number of ‘permutations’ is $6$ but the number of ‘combinations’ is just $1$.

The problem becomes more complex when only a few of the items are considered for the arrangements. Also it may be long exercise if you apply the counting principle in most cases. Instead, there are ready to use formulas derived by mathematicians as described below.

If there are ‘$n$’ items when only ‘$r$’ items are considered and if $s(1), s(2), ….s(k)$ are ‘$k$’ sets of similar items, then the number of permutation is denoted and given by,

$^nP_r$ = $\frac{n!}{(n - r)!\ \times\ s(1)!\ \times\ s(2)!….s(k)!}$

The formula for combination of ‘$n$’ items with ‘$r$’ items taken at a time is,

$^nC_r$ = $\frac{n!}{((n-r)!\ \times\ r!}$

With the help of these formulas we can determine the number of all possible events in a situation which helps us to find the probability of a desired event to happen. Thus the concept of permutations and combinations helps us in determining the probability in certain cases.

## Word Problems

Example 1:

A packet contains similar size of cards labeled as $A, B, C, D$ and $E$. Three cards are picked up one by one at random and placed on the same order. What is the probability that the word formed by those letters is $BAD$?

Solution:

Since the number of items (the letters)is small, we can use the principle of counting to ascertain the total number of three letter words (with or without meaning) can be formed.

The first letter of the word can be selected from any of the given $5$ letters. Hence there are five number of ways by which the first letter is selected. Then the second letter could be selected only in $4$ ways and the third in $3$ ways. Therefore, the total number of all possible three letter words (with or without meaning) is $5 \times 4 \times 3$ = $60$. Even if you use the permutation formula, we get the same number.

This is nothing but the sample space or the total number of all possible outcomes for the given probability situation. The desirable outcome is only $1$, namely the letter appearing $B$ as first, the letter $A$ as second and the letter $C$ as third. Therefore, the probability for the word $BAD$ to be formed is, $\frac{1}{60}$.
Example 2:

A computer program allots a $4$ digit OTP to the customers by randomly selection from the digits $0, 1, 2, 3, 4, 5$ and $6$. The system does generate repeated digits. What is the probability that a customer gets $1234$ as the OTP?

Solution:

Since the first place cannot have the digit $0$, it can be occupied only from the remaining $5$ numbers. But, for the second digit $0$ also could be selected in $5$ ways. But the third and fourth places can be selected only in $4$ and $3$ ways respectively. Hence the total number of permutations (the sample space) is $5 \times 5 \times 4 \times 3$ = $300$. Therefore the probability of getting $1234$ as OTP is $\frac{1}{300}$.
Example 3:

A committee of $4$ persons are to be formed from the club members consisting $20$ men and $15$ women. What is the probability of a committee consisting $2$ men and $2$ women is formed?

Solution:

This is a case of combination since there is no order to be followed. It is just the combinations of $3$ persons out of $35$ persons. Hence the total number of ways of forming a committee of $3$ persons is, $^{35}C_4$ = $\frac{35!}{(31)! \times 4!}$ = $52360$. This is the sample space.

The number of ways by which a committee of $2$ man and $2$ women formed is a question of combination of $2$ men out of $20$ men times a combination of $2$ women out of $15$ men. That is,

$^{20}C_2 \times ^{15}C_2$ = $\frac{20!}{(18)! \times 2!}$ $\times$ $\frac{15!}{(13)! \times 2!}$ = $190 \times 105$ = $19950$ and this is the number of favorable outcomes.

Hence, the required probability = $\frac{19950}{52360}$ $\approx\ 0.38$
Example 4:

Four cards are drawn from $s$ standard deck of cards. What is the probability that all the four cards are of same suit.

Solution:

This is also a case of combination since no order is to be followed. Firstly, the total number of ways that $4$ cards can be drawn out of $52$ cards is,

$^{52}C_4$ = $\frac{52!}{(48)!\ \times\ 4!}$ = $270725$. This is the sample space.

The number of ways by which 4 cards of the same suit are drawn is,

$^{13}C_4\ +\ ^{13}C_4\ +\ ^{13}C_4\ +\ ^{13}C_4$ = $4\ \times$ $\frac{13!}{(9)! \times 4!}$ = $2860$ and this the number of favorable outcomes.

Hence, the required probability = $\frac{2860}{270725}$ $\approx\ 0.011$