Permutation & Combinations Concept


Factorial Notation:
Let n be positive integer.Then, factorial n denoted by n!
is defined as n! = n(n-1)(n-2). . . .  . . .  .3.2.1
             eg:- 5! = (5 * 4* 3 * 2 * 1)
              = 120
              0! = 1


Permutations:
The different arrangements of a given number of things by
taking some or all at a time,are called permutations.
eg:- All permutations( or arrangements)made with the letters
a,b,c by taking two at a time are (ab, ba, ac, ca, bc, cb)

                          Permutations
Numbers of permutations:
Number of all permutations of n things, taken r at a time is
given by  nPr  = n(n-1)(n-2). .  .. . . (n-r+1)
               = n! / (n-r)!

An Important Result:
If there are n objects 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!)

Combinations:
Each of different groups or selections which can be formed by
taking some or all of a number of objects,is called a combination.
  eg:- Suppose we want to select two out of three boys A, B, C.
         then, possible selection are AB, BC & CA.
         Note that AB and BA represent the same selection.

Number of Combination:
The number of all combination of n things taken r at a time is:
    nCr  = n! / (r!)(n-r)!
     = n(n-1)(n-2). . . . . . . tor factors / r!
    Note: nCn = 1 and nC0 =1
   An Important Result:
   nCr = nC(n-r)
                               Permutation & Combinations                 
Problems:

1.Evaluate 30!/28!

Sol:-      30!/28! = 30 * 29 * (28!)  / (28!)
                   = 30 * 29 =870

2.Find the value of 60P3

Sol:- 60P3  = 60! / (60-3)! = 60! / 57!
            = (60 * 59 *58 * (57!) )/ 57!
            = 60 * 59 *58
            = 205320

3. Find the value of 100C98,50C 50

Sol:-       100C98   = 100C100-98)
                     = 100 * 99 / 2 *1
                    = 4950
              50C50 = 1

4.How many words can be formed by using all the letters of the
word DAUGHTER so that vowels always come together & vowels are never together?

Sol:-
 (i) Given word contains 8 different letters
     When the vowels AUE are always together we may suppose
      them to form an entity, treated as one letter
      then the letter to be arranged are DAHTR(AUE)
      these 6 letters can be arranged in 6p6 = 6!
       = 720 ways
  The vowels in the group (AUE) may be arranged in 3! = 6 ways
            Required number of words = 760 * 6 =4320


(ii) Total number of words formed by using all the letters of
the given words
         
      8! = 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1
         = 40320
   Number of words each having vowels together
                  = 760 * 6
                 = 4320
  Number of words each having vowels never together
               = 40320-4320  
              = 36000

5.In how many ways can a cricket eleven be chosen out of a batch
of 15 players.

Sol:- Required number of ways
                          = 15C 11 = 15C (15-11)
                          = 15 C 4
          15C4 = 15 * 14 * 13 * 12 / 4 * 3 * 2 *1
              = 1365

6.In how many  a committee of 5 members can be selected from 6 men
5. ladies consisting of 3 men and 2 ladies

Sol:-   (3 men out of 6) and (2 ladies out of 5) are to be chosen
              Required number of ways
                        =(6C3 * 5C2)
                        = 200

7.How many 4-letter word with or without meaning can be formed out
of the letters of the word 'LOGARITHMS' if repetition of letters is
not allowed

Sol:-        'LOGARITHMS' contains 10 different letters
                Required number of words
             = Number of arrangements of 100 letters taking
               4 at a time
             = 10 P4
             = 10 * 9 * 8 * 7
             = 5040

8.In how many ways can the letter of word 'LEADER' be arranged

Sol:-     The word 'LEADER' contains 6 letters namely
                           1L,2E,1A,1D

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