$n! = n(n-1) (n-2) ......3.2.1$

$5! = 5 \times 4 \times 3 \times 2 \times 1 =120$

Note $0! =1$

$n! =n \times (n-1)!$

or $(n-1)! = \frac {n!} {n}$

Putting n = 1, we have

$0! = \frac {1!}{1}$

or $ 0! = 1$

$(n+2)!=2550 \times n!$

$(n+2)!=2550 \times n!$

$(n+2)(n+1)n!=2550 \times n!$

Canceling n!

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

$n=49$

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?

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

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?

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 $(20 \times 30)=600$ ways

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

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

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:

$^{n}P_r = n (n-1)(n-2).....(n-r+1)$

$= \frac {[n(n-1)(n-2)...(n-r+1)] [(n-r)(n-r-1)...3.2.1.]}{ [(n-r)(n-r-1)...3.2.1]}$

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

$^{n}P_n= n!$

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

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

$^{n}P_n= n!$

$0! = 1$

We have $^{n}P_r=\frac { n!}{ (n-r)!}$

Putting r = n, we have :-

$^{n}P_n =\frac { n!}{ (n-n)!}= \frac {n!}{0!}$

But $^{n}P_n= n!$

Clearly it is possible, only when $0! = 1$

Hence it is proof that $0! = 1$

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

Number of ways taking 4 flags out of 6-flags = $^{6}P_4=\frac { 6!}{ (6-4)!}$

$=6 \times 5 \times 4 \times 3 = 360$

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

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

Number of Permutations taking all the letters at a time = $^{n}P_n$

$=5! = 5 \times 4 \times 3 \times 2 \times 1=120$

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

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

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

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

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

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

such that $(p_1 + p_2 + ... p_r) = n$.

Then, number of permutations of these n objects is =$\frac {n!}{p_1! p_2!....p_r!}$

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

- T’ and ‘R’ occupying end places.
- G’ being always in the middle
- Vowels occupying odd-places
- Vowels being never together.
- Find the number of words formed with out any restriction

- 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 =$3! \times 2! = 12$ ways. - 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. - 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= $3! \times 3! = 36$ ways. - 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 = $4! \times 2!= 48$ ways

Number of ways, when vowels being never-together

$= 120-48 =72$ ways. - Total number of words = 5! = 120

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

- INDIA
- RUSSIA
- AUGUST
- ARRANGE

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

So Number of words formed= $\frac {n!}{p_1! p_2!.... p_r!}$

$=\frac {5!}{2!} =60$ words - In RUSSIA, There are 2 S’s and all other are distinct

So Number of words formed= $\frac {n!}{p_1! p_2!.... p_r!}$

$ =\frac {6!}{2!} =360$ words - In AUGUST, There are 2 U’s and all other are distinct

So Number of words formed= $\frac {n!}{p_1! p_2!.... p_r!}$

$ =\frac {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= $\frac {n!}{p_1! p_2!.... p_r!}$

$ =\frac {7!}{2! \times 2!}= =1260$ words

- Circular permutation refers to the arrangement of objects in a circle, where the order of the objects matters but rotations of the arrangement are considered identical. In other words, rotating the arrangement does not create a new permutation. For example, in a circular permutation of three objects A, B, and C, the arrangements ABC, BCA, and CAB are considered the same.
- The number of distinct circular permutations of n objects is given by (n−1)!. This formula accounts for the fact that fixing one object as a reference point leaves
- The number of circular permutations of n different objects is (n-1)!/2 If anti-clockwise and clockwise order of arrangement are same as in case of beads
- Similary Number of circular permutations of n different things, taken r at a time, when clockwise and anticlockwise orders are taken as different
$\frac {^{n}P_{r}}{r}$

- Number of circular permutations of n different things, taken r at a time, when clockwise and anticlockwise orders are taken are same
$\frac {^{n}P_{r}}{2r}$

**Notes**- What is factorial
- Fundamental Principle of counting
- Permutation
- Restricted Permutations
- Another Theorem of Permutation
- What is combination
- Combination Formula
- Properties of Combination Formula

**NCERT Solutions**