- What is factorial
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- Fundamental Principle of counting
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- Permutation<
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- Restricted Permutations
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- Another Theorem of Permutation

The product of first ‘n’ natural numbers is denoted by n!.

n! = n(n-1) (n-2) ………………..3.2.1.

**Example**

5! = 5 x 4 x 3 x 2 x 1 =120

Note 0! = 1

Proof n! =n, (n-1)!

Or (n-1)! = [n x (n-1)!]/n = n! /n

Putting n = 1, we have

0! = 1!/1

or 0! = 1

**Example**

(n+2)!=2550 X n!

Solution:

(n+2)!=2550 X n!

(n+2)(n+1)n!=2550 X n!

Canceling n!

(n+2)(n+1)=2550

N=49

**Addition rule **: If an job can be performed in ‘n’ ways, & another job can be performed in ‘m’ ways then either of the two jobs can be performed in (m+n) ways. This rule can be extended to any finite number of jobs.

**Example**: In a class 12 , 20 students are from Maths streams while 30 students are from Biology stream.

Teacher has to choose 1 student either from Maths stream or Biology stream to represent the class

What ways the selection can be made?

**Solution:**

20 students are from Maths:

Students can be chosen from Maths stream in 20 ways

30 students are from Biology stream

Students can be chosen from Biology stream in 30 ways

So by the Fundamental Principle of addition rule, the selection of 1 student from either of two streams can be done (20+30)=50 ways

**Multiplication Rule **: If a work can be done in m ways, another work can be done in ‘n’ ways, then both of the operations can be performed in m x n ways. It can be extended to any finite number of operations

**Example:** In a class 12 , 20 students are from Maths streams while 30 students are from Biology stream.

Teacher has to choose 2 student one from Maths stream and one from Biology stream to represent the class in Quiz competition

What ways the selection can be made?

**Solution:**

20 students are from Maths:

1 Students can be chosen from Maths stream in 20 ways

30 students are from Biology stream

1Students can be chosen from Biology stream in 30 ways

So by the Fundamental Principle of Multiplication rule, the selection of 1 student from Maths streams and 1 students from Biology stream can be done (20X30)=600 ways

Permutation means arrangement of things. The word arrangement is used, if the order of things is considered

**Theorem 1:**

Number of permutations of ‘n’ different things taken ‘r’ at a time is given by:-

^{n}P_{r} = n! / (n-r)!

**Proof:** Say we have ‘n’ different things a_{1} , a_{2} …… , a_{n}

- Clearly the first place can be filled up in ‘n’ ways. Number of things left after filling-up the first place = n-1
- So the second-place can be filled-up in (n-1) ways. Now number of things left after filling-up the first and second places = n – 2
- Now the third place can be filled-up in (n-2) ways.
Thus number of ways of filling-up first-place = n

Number of ways of filling-up second-place = n-1

Number of ways of filling-up third-place = n-2

Number of ways of filling-up r-th place = n – (r-1) = n-r+1

By multiplication – rule of counting, total no. of ways of filling up, first, second -- rth-place together :-

n (n-1) (n-2) ------------ (n-r+1)

Hence:

^{n}P_{r}= n (n-1)(n-2) --------------(n-r+1)= [n(n-1)(n-2)----------(n-r+1)] [(n-r)(n-r-1)-----3.2.1.] / [(n-r)(n-r-1)] ----3.2.1

**nPr = n!/(n-r)!**

**Theorem 2:**Number of permutations of ‘n’ different things taken all at a time is given by:-

^{n}P_{n} = n!

**Proof :**

Now we have ‘n’ objects, and n-places.

Number of ways of filling-up first-place = n

Number of ways of filling-up second-place = n-1

Number of ways of filling-up third-place = n-2

Number of ways of filling-up r-th place, i.e. last place =1

Number of ways of filling-up first, second, --- n th place

= n (n-1) (n-2) ------ 2.1.

^{n}P_{n}= n!

**To prove**

0! = 1

We have

**nPr = n!/(n-r)!**

Putting r = n, we have :-

^{n}P_{n} = n! / (n-n)!= n!/0!

But ^{n}P_{n} = n!

Clearly it is possible, only when 0! = 1

Hence it is proof that 0! = 1

**Example** How many different signals can be made by 4 flags from 6-flags of different colours?

**Solution :**

Number of ways taking 4 flags out of 6-flags = ^{6}P_{4} = 6!/(6-4)!

= 6 x 5 x 4 x 3 = 360

**Example** How many words can be made by using the letters of the word “ZEBRA” taken all at a time?

**Solution** . There are ‘5’ different letters of the word “ZEBRA”

Number of Permutations taking all the letters at a time = ^{5}P_{5}

= 5! = 5X4X3X2X1=120

**Theorem ** Number of permutations of ‘n’ things, taken ‘r’ at a time, when a particular thing is to be always included in each arrangement

= r ^{n-1}P_{r-1}

**Proof:**

When one particular thing is always taken, then that can arranged in r ways

Now we have to arrange the remaining r-1 places with n-1 objects,so permutations will be ^{n-1}P_{r-1} . Now by principle of counting ,both of these can be arrange in r ^{n-1}P_{r-1}

**Theorem:** Number of permutations of ‘n’ things, taken ‘r’ at a time, when a particular thing is fixed:

= ^{n-1}P_{r-1}

**Proof:**

When one particular thing is fixed ,then that is the only arrangement for that.Now we have to arrange the remaining r-1 places with n-1 objects,so permutations will be ^{n-1}P_{r-1} . Now by principle of counting ,both of these can be arrange in ^{n-1}P_{r-1}

**Theorem:** Number of permutations of ‘n’ things, taken ‘r’ at a time, when a particular thing is never taken

= ^{n-1}P_{r}

**Proof:**

When one particular thing is never taken then we have to arrange the r- places with n-1 objects,so permutations will be ^{n-1}P_{r} .

**Theorem**: There are n subjects of which p1 are alike of one kind; p2 are alike of another kind; p3 are alike of third kind and so on and pr are alike of rth kind,

such that (p1 + p2 + ... pr) = n.

Then, number of permutations of these n objects is =n!/p1! p2!.... pr!

**Example: **How many words can be formed with the letters of the word ‘TIGER’ when:

(i) ‘T’ and ‘R’ occupying end places.

(ii) ‘G’ being always in the middle

(iii) Vowels occupying odd-places

(iv) Vowels being never together.

v) Find the number of words formed with out any restriction

**Solution**

(i) When ‘T’ and ‘R’ occupying end-places

IGE. (TR)

Here (TR) are fixed, hence I,E,G can be arranged in 3! ways

But (T,R) can be arranged themselves is 2! ways.

Total number of words = 3! x 2! = 12 ways.

(ii) When ‘G’ is fixed in the middle

T.I.(G), E.R

Hence four-letter T.I.E.R. can be arranged in 4! i.e 24 ways.

(iii) Two vowels (I,E) can be arranged in the odd-places (1st, 3rd and 5th) = 3! ways.

And three consonants (T,G,R) can be arranged in the remaining places = 3 ! ways

Total number of ways= 3! x 3! = 36 ways.

(iv) Total number of words = 5! = 120

If all the vowels come together, then we have: (I.E.), T,R,G

These can be arranged in 4! ways.

But (I,E.) can be arranged themselves in 2! ways.

Number of ways, when vowels come-together = 4! x 2!

= 48 ways

Number of ways, when vowels being never-together

= 120-48 = 72 ways.

v) Total number of words = 5! = 120

**Example:**

Find the numbers of words formed by permuting all the letters of the following words

1) INDIA

2) RUSSIA

3) AUGUST

4) ARRANGE

**Solution**

1) In INDIA, There are 2 I’s and all other are distinct

So Number of words formed= n!/p1! p2!.... pr!

=5!/2! =60 words

2) In RUSSIA, There are 2 S’s and all other are distinct

So Number of words formed= n!/p1! p2!.... pr!

=6!/2! =360 words

3) In AUGUST, There are 2 U’s and all other are distinct

So Number of words formed= n!/p1! p2!.... pr!

=6!/2! =360 words

4) In ARRANGE, There are 2 R’s and 2A’s and all other are distinct

So Number of words formed= n!/p1! p2!.... pr!

=7!/2!. 2!= =1260 words

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