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

n! = n(n-1) (n-2) ………………..3.2.1.

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

(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

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

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 (20X30)=600 ways

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

- 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)!**

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.

0! = 1

We have

Putting r = n, we have :-

But

Clearly it is possible, only when 0! = 1

Hence it is proof that 0! = 1

Number of ways taking 4 flags out of 6-flags =

Number of Permutations taking all the letters at a time =

= 5! = 5X4X3X2X1=120

= r

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

=

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

=

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

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

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

(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

(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

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

1) INDIA

2) RUSSIA

3) AUGUST

4) ARRANGE

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

Go Back to Class 11 Maths Home page Go Back to Class 11 Physics Home page

- Mathematics - Class 11 by RD Sharma
- NCERT Exemplar Problems: Solutions Mathematics Class 11
- NCERT Solutions: Mathematics Class 11th
- Mathematics Textbook for Class XI
- Cbse Mathematics for Class XI (Thoroughly Revised As Per New Cbse Syllabus)
- 37 Years Chapterwise Solved Papers (2015-1979): IIT JEE - Mathematics
- Play with Graphs - Skills in Mathematics for JEE Main and Advanced