Permutation

for the permutation we first need to understand factorial. 

n!(read as n factorial)=n(n-1)!=n(n-1)(n-2)!= n(n-1)(n-2)(n-3)!

in simple words n factorial is the multiplication of all previous numbers.

for example:
n=5

n!=5!=5*4*3*2*1=120

rules of factorial:

0!=1

1!=1

Permutation

Permutation is an arrangement of things or characters or digits or anything which we have. In simple words, permutation is a number of arrangements that a number of things can have.

Formula to find permutations:

P(n,r) = npr = n!/(n-r)!   { Here, n!=n(n-1)!=n(n-1)(n-2)!=n(n-1)(n-2)(n-3)!}

rules to calculate permutation:

P(n,0)=nPo=1

P(n,1)=nP1=n

P(n,n-1)=nP(n-1)=n

For example:

What are the different arrangements possible for the first 5 digits when we need only 4 digits at a time.

5P4=5!/4!=(5×4!)/(4!)=5

1.How many words can be made from the first 10 alphabets(capital) of lengh 6?

solution:

Here n=10

r=6

Total words from the alphabets are:

10P6=10!/6!=10×9×8×7×6!/7!=

How many permutations are possible for 4 alphabets?

Here, 

It is simply asked to find different arrangements of the 4 alphabets.

=4!

=4×3×2×1

=24 permutations 

2. How many words (meaningful or otherwise) of length 3 can be formed from the word "COMPUTER", the beginning letter be C and no repetition.

solution:

the given word contains  8 distinct letters. Since the first letter is fixed, the remaining 2 letters can be :

7*6 = 42 ways.

3. Find how many symbol codes can be formed if the first if the first two symbols are letters and the next three are digits, but no symbol is repeated.

solution:
here we have to fill 5 positions in which 1st  two position can be filled by P(26, 2){total number of alphabets are 26 and we have to choose 2 of them} and the next three positions are filled by P(10,3) {as the total numbers of digits from 0 to 9 i.e. 10 and we have to take 3 }

total Permutations = (26!/(26-2)*10!/(10-3)) = ((26!/24!)*(10!/7!))=26*25*10*9*8=468000.

4.  (i)suppose repetitions are not allowed, then how many 4 digit numbers can be formed from the six digits 1,2,3,5,7,8?

(ii) How many such numbers are less than 4000?

(iii) How many of numbers of (i) are even?

(iv) How many of numbers in (ii) are odd?

(v) How many of the numbers in (i) contain both the digits 3 and 5? 

solution:

(i)  the number of four digits is given by P(6,4) = 6!/2! = 360

(ii) the number beginning with 1,2,and 3 are less than 4000. the number of integers beginning with 1 are : 5*4*3 = 60

the number beginning with 2: 5*4*3 = 60

the number beginning with 3: 5*4*3 = 60

thus the total number of integers less than 4000 is 60+60+60 = 180

(iii) the numbers which are ending with 2 or 8 is even. Numbers ending in 2 are 5*4*3 = 60. similarly, the numbers starting with 8 are: 5*4*3 = 60. hence the number which are even are: 60+60=120.

(iv) the numbers ending in 1,3,5 and 7 are odd. the 4 digit numbers ending in 1 and less than 4000 are begin with either 2 or 3. so these total ways are: 2*4*3=24.

similarly the numbers ending in 3 are 24.

however the numbers ending in 5 or 7 are 2*3*4*3=72. 

total are= 2*24+72=120 such numbers.

(v) the digit 3 can occupy any of the 4 positions and the remaining 3 are occupied by digit 5. hence the number of ways in which the digit 3 and 5 can appear is 4*3 = 12. Hence the number of integers containing both 3 and 5 is 12*4*3 = 144

5. how many different ways are there to arrange the letters in the word "PROBLEM" if 

(i) the letters P must first.

(ii) the letter P must come first and the letter M last?

solution:

(i) since the letter P must come first, its position is fixed. Hence the remaining 6 letters can be arranged in 6! ways. Hence the required answer is 6!.

(ii) since the letter P must be come first and the letter M last, the remaining 5 letters can be arranged in 5! ways.