Mathematics PERMUTATIONS - BASICS AND DEFINITION
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### Topics Covered

star Permutations
star Permutations when all the objects are distinct
star Factorial notation

### Permutations

if we have to determine the number of 3-letter words, color(blue)"with" or color(blue)"without meaning", which can be formed out of the letters of the word color(red)"NUMBER,"

Where color(blue)"the repetition of the letters is not allowed", we need to count the arrangements "NUM, NMU, MUN, NUB, ..., " etc. Here, we are counting the permutations of 6 different letters taken 3 at a time. The required number of words color(red)(= 6 × 5 × 4 = 120) "(by using multiplication principle)."

color(red)"Definition " A permutation is an arrangement in a definite order of a number of objects taken some or all at a time.

### Permutations when all the objects are distinct

color(red)("Theorem 1 ") The number of permutations of n different objects taken r at a time, where color(blue)(0 < r ≤ n) and the objects do not repeat is color(red)(n* ( n – 1)* ( n – 2)..... ( n – r + 1),) which is denoted by color(red)(text()^nP_r.)

color(red)"Proof " There will be as many permutations as there are ways of filling in r vacant places color(blue)(underset (leftarrow \ " r vacant places " ->) (square \ square \ square cdots \ square)) by the n objects.

The first place can be filled in n ways; following which, the second place can be filled in (n – 1) ways, following which the third place can be filled in (n – 2) ways,..., the r^(th) place can be filled in (n – (r – 1)) ways.

Therefore, the number of ways of filling in r vacant places in succession is n(n – 1) (n – 2) . . . (n – (r – 1)) or n ( n – 1) (n – 2) ... (n – r + 1).

### Factorial notation

The notation color(red)(n!) represents the product of first n natural numbers, i.e., the product 1 × 2 × 3 × . . . × (n – 1) × n is denoted as n!. We read this symbol as ‘n factorial’. Thus, color(red)(1 × 2 × 3 × 4 . . . × (n – 1) × n = n !)
color(blue)(1 = 1 !)

\color{green} ✍️ color(blue)(1 × 2 = 2 !)

\color{green} ✍️ color(blue)(1× 2 × 3 = 3 !)

1 × 2 × 3 × 4 = 4 !  and so on.

\color{green} ★\color{green} \mathbf(KEY \ CONCEPT)

We define color(red)(0 ! = 1)

We can write 5 ! = 5 × 4 ! = 5 × 4 × 3 ! = 5 × 4 × 3 × 2 ! = 5 × 4 × 3 × 2 × 1!

Clearly, for a natural number n
\color{green} ✍️ color(red)( n ! = n (n – 1) !)

= n (n – 1) (n – 2) ! \ \ \ \ \ \ [provided (n ≥ 2)]

= n (n – 1) (n – 2) (n – 3) ! \ \ \ \ \ \ [provided (n ≥ 3)]

and so on.
Q 3078234106

Evaluate (i) 5 ! (ii) 7 ! (iii) 7 ! – 5!

Solution:

(i) 5 ! = 1 × 2 × 3 × 4 × 5 = 120
(ii) 7 ! = 1 × 2 × 3 × 4 × 5 × 6 ×7 = 5040 and (iii) 7 ! – 5! = 5040 – 120 = 4920.
Q 3038434302

Compute (i). (7!)/(5!)

(ii). (12 !)/{(10!)(2!)}

Solution:

(i) We have (7!)/(5!) = (7xx6xx5!)/(5!) = 7xx6 = 42

(ii) (12 !)/{(10!)(2!)} = (12xx11xx(10!) )/{(10!)(2)} = 6xx11 = 66.
Q 3088534407

Evaluate (n!)/{ r! (n-r)!} when n = 5, r = 2.

Solution:

We have to evaluate (5!)/{2! (5-2)!} (since n = 5, r = 2)

We have (5!)/{2! (5-2)!} = (5!)/(2! xx 3!) = (4xx5)/2 = 10
Q 3078634506

If 1/(8!) +1/(9!) = x/(10!) find x.

Solution:

We have 1/(8!) +1/(9!) = x/(10!)

1/(8!) +1/(9 xx8 !) = x/(10xx9xx8!)

(1+1/9) = x/(10xx9)

10/9 = x/(10xx9)

so x = 100