In the previous section, we completed a discussion on the multiplication principle. In this section, we will see Permutations.
Some basics about permutations can be understood by considering the example that we saw in a previous section. It can be written in 3 steps:
1. In the example 7.1 in the first section of this chapter, we arranged the four letters of the word ROSE in different possible ways.
• We saw that, the letters can be arranged in 24 different ways.
♦ So we have a list of 24 words.
♦ Each of the 24 words in that list is unique.
2. This uniqueness can be explained in 3 steps:
(i) Let us consider one of the word in the list, say SREO
(ii) In this word,
♦ S is in the first position
♦ R is in the second position
♦ E is in the third position
♦ O is in the fourth position
• We must not change this order.
(iii) If we change the order, we will get a ‘different word’.
♦ This ‘different word’ will be already present in the list.
♦ Then we cannot say that, the list contains 24 different words.
♦ Because one of them is repeating.
♦ The list will then become invalid.
• That is why, we say that, each member of the list is unique. We must not alter the arrangement in that member.
3. Each arrangement is called a permutation of 4 different letters taken all at a time.
◼ We can write:
There are 24 permutations.
Let us see another example. It can be written in 3 steps:
1. Consider the word NUMBER.
How many 3 letter words, with or without meaning, can be formed using the letters in this word. Repetition of letters is not allowed.
Answer: There are 6 letters in the word NUMBER. So applying the multiplication principle, the answer is (6 × 5 × 4) = 120
♦ So we have a list of 120 words.
✰ Some examples are: NBR, MRU, RNE etc.,
♦ Each of the 120 words in that list is unique.
2. This uniqueness can be explained in 3 steps:
(i) Let us consider one of the word in the list, say RNE
(ii) In this word,
♦ R is in the first position
♦ N is in the second position
♦ E is in the third position
• We must not change this order.
(iii) If we change the order, we will get a ‘different word’.
♦ This ‘different word’ will be already present in the list.
♦ Then we cannot say that, the list contains 120 different words.
♦ Because one of them is repeating.
♦ The list will then become invalid.
• That is why, we say that, each member of the list is unique. We must not alter the arrangement in that member.
3. Each arrangement is called a permutation of 6 different letters taken 3 at a time.
◼ We can write:
There are 120 permutations.
Now we can write the definition for permutation. It can be written in 3 steps:
(i) A permutation is an arrangement of objects in a definite order.
(ii) In some problems, all available objects will be included in each of the arrangements.
(iii) In some problems, only a certain number of the available objects will be included in each of the arrangements.
Next we will try to derive formulas for various types of problems.
A. Formula for the number of permutations of n objects taken r at a time
This formula can be derived in 3 steps:
1. Suppose that there are a total of n objects.
• If we take r objects at a time and try the permutations, then the number of permutations possible is given by: n(n-1)(n-2)(n-3) . . . (n-r+1)
2. The proof can be written in 2 steps:
(i) If we take r objects at a time, there will be r boxes
• The first box can be filled in n different ways
• The second box can be filled in (n-1) different ways
♦ (n-1) is [n-(2-1)]
♦ So for second box, we have the term (2-1)
• The third box can be filled in (n-2) different ways
♦ (n-2) is [n-(3-1)]
♦ So for third box, we have the term (3-1)
• The fourth box can be filled in (n-3) different ways
♦ (n-3) is [n-(4-1)]
♦ So for fourth box, we have the term (4-1)
- - -
- - -
• Based on the pattern, we get:
The rth box can be filled in [n-(r-1)] different ways
⇒ The rth box can be filled in (n-r+1) different ways.
(ii) So by applying the multiplication principle, we get:
Total number of arrangements (permutations) = n(n-1)(n-2)(n-3) . . . (n-r+1)
3. The expression n(n-1)(n-2)(n-3) . . . (n-r+1) is denoted as nPr
◼ So we can write the formula:
The number of permutations of n objects taken r at a time is given by:
nPr = n(n-1)(n-2)(n-3) . . . (n-r+1)
◼ Here two points are to be noted
First point can be written in 4 steps:
(i) n is a number of objects. So it has to be greater than 0
♦ That is., 0 < n
(ii) r is a number of objects. So it has to be greater than 0
♦ That is., 0 < r
(iii) We are taking r objects out of n objects. So r must be less than or equal to n
♦ That is., r ≤ n
(iv) Combining the three inequalities, we get: 0 < r ≤ n
Second point can be written in 3 steps:
(i) The n objects must be different
(ii) For example, suppose that the objects are the letters of the word NUMBER.
Here n is 6. All 6 objects are different.
(iii) Suppose that the objects are the letters of the word ROOTS
Here n is 5. All 5 objects are not different because, the letter ‘O’ appears two times.
• So we obtained an expression for calculating the number of permutations:
nPr = n(n-1)(n-2)(n-3) . . . (n-r+1)
• This expression is cumbersome. We must try to shorten it.
• The symbol n! can help us to shorten the expression.
• So our next task is to learn about n!. It can be written in steps:
1. n! is read as factorial n. It can also be read as n factorial.
2. When we write n!, the n must be a natural number.
3. n! is the product of all natural numbers beginning from 1 and ending at n
• So we can write:
n! = 1 × 2 × 3 × 4 × . . . × (n-1) × n
• Let us see some examples:
1! = 1
2! = 2 × 1
3! = 3 × 2 × 1
4! = 4 × 3 × 2 × 1
5! = 5 × 4 × 3 × 2 × 1
- - -
- - -
so on . . .
4. Mathematicians define 0! as 1
♦ That is., 0! = 1
• We will see the proof in later sections.
5. Now we will see some interesting results. It can be written in 4 steps:
(i) We know that, 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1
• This can be written as: 7! = 7 × [6 × 5 × 4 × 3 × 2 × 1]
♦ But [6 × 5 × 4 × 3 × 2 × 1] is 6!
• So we get: 7! = 7 × 6!
• Based on this example, we can write: n!= n × (n-1)!
(ii) We know that, 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1
• This can be written as: 7! = 7 × 6 × [5 × 4 × 3 × 2 × 1]
♦ But [5 × 4 × 3 × 2 × 1] is 5!
• So we get: 7! = 7 × 6 × 5!
• Based on this example, we can write: n!= n × (n-1) × (n-2)!
(Here n must be greater than or equal to 2. Otherwise, (n-2) will become -ve.)
(iii) We know that, 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1
• This can be written as: 7! = 7 × 6 × 5 × [4 × 3 × 2 × 1]
♦ But [4 × 3 × 2 × 1] is 4!
• So we get: 7! = 7 × 6 × 5 × 4!
• Based on this example, we can write: n!= n × (n-1) × (n-2) × (n-3)!
(Here n must be greater than or equal to 3. Otherwise, (n-3) will become -ve.)
(iv) So we can write a general form:
n!= n × (n-1)!
n!= n × (n-1) × (n-2)! [provided n ≥ 2]
n!= n × (n-1) × (n-2) × (n-3)! [provided n ≥ 3]
- - -
- - -
so on . . .
Now we will see some solved examples
Solved example 7.6
Evaluate (i) 5! (ii) 7! (iii) 7! - 5!
Solution:
Part (i):
5! = 5 × 4 × 3 × 2 × 1 = 120
Part (ii):
7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040
Alternate method:
7! = 7 × 6 × 5! = 42 × 120 = 5040
Part (iii):
7! - 5! = (5040 - 120) = 4920
Solved example 7.7
Evaluate (i) $\frac{7!}{5!}$ (ii) $\frac{12!}{(10!)(2!)}$
Solution:
Part (i):
$\begin{array}{ll}
{}&\frac{7!}{5!} &{}={}& \frac{7 × 6 × 5!}{5!} &{} \\
{}&\phantom{\frac{7!}{5!}}&{}={}&7 × 6 \\
{}&\phantom{\frac{7!}{5!}}&{}={}&42 \\
\end{array}$
Part (ii):
$\begin{array}{ll}
{}&\frac{12!}{(10!)(2!)} &{}={}& \frac{12 × 11 × 10!}{(10!)(2 × 1)} &{} \\
{}&\phantom{\frac{12!}{(10!)(2!)}}&{}={}&\frac{12 × 11}{(2 × 1)} \\
{}&\phantom{\frac{12!}{(10!)(2!)}}&{}={}&6 × 11 \\
{}&\phantom{\frac{12!}{(10!)(2!)}}&{}={}&66 \\
\end{array}$
Solved example 7.8
Evaluate $\frac{n!}{r! (n-r)!}$ when n = 5, r = 2
Solution:
$\begin{array}{ll}
{}&\frac{n!}{r! (n-r)!} &{}={}& \frac{5!}{2! (5-2)!}\\
{}&\phantom{\frac{n!}{r! (n-r)!}}&{}={}&\frac{5 × 4 × 3!}{2! (3)!} \\
{}&\phantom{\frac{n!}{r! (n-r)!}}&{}={}&\frac{5 × 4}{2 × 1} \\
{}&\phantom{\frac{n!}{r! (n-r)!}}&{}={}&5 × 2 \\
{}&\phantom{\frac{n!}{r! (n-r)!}}&{}={}&10 \\
\end{array}$
Solved example 7.9
$\rm{If}~\frac{1}{8!}+\frac{1}{9!}=\frac{x}{10!},~\rm{find x}$
Solution:
$\begin{array}{ll}
{}&\frac{1}{8!}+\frac{1}{9!} &{}={}& \frac{x}{10!}&{} \\
{\Rightarrow}&\frac{1}{8!}+\frac{1}{9!}&{}={}& \frac{x}{10 × 9 × 8!}&{} \\
{\Rightarrow}&\frac{10
× 9 × 8!}{8!}+\frac{10 × 9 × 8!}{9!}&{}={}& \frac{x × 10 × 9 ×
8!}{10 × 9 × 8!}&{\color {green}{\text{(Multiplying by 10 × 9 ×
8!)}}} \\
{\Rightarrow}&\frac{10 × 9 × 8!}{8!}+\frac{10 × 9 ×
8!}{9 × 8!}&{}={}& \frac{x × 10 × 9 × 8!}{10 × 9 × 8!}&{} \\
{\Rightarrow}&10 × 9+10&{}={}& x&{} \\
{\Rightarrow}&90+10&{}={}& x&{} \\
{\Rightarrow}&100&{}={}& x&{} \\
\end{array}$
The link below gives some more solved examples:
Now we have a basic idea about factorial notation. Remember that, we are trying to shorten the expression for nPr. In the next section, we will see how the factorial notation will help us to achieve this.
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