`star` Permutations

`star` Permutations when all the objects are distinct

`star` Factorial notation

`star` Permutations when all the objects are distinct

`star` Factorial notation

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.

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.

`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).`

`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).`

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.

`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!`

(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!)}`

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

(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.`

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`.

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`