PERMUTATIONS & COMBINATIONS

Formulae: 
 
Factorial Notation:
Let n be positive integer. Then ,factorial n dentoed 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)

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 atime 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)
                                                
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 DAUGHTR 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 6men
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
             = 10P4
             = 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 and 1R
               Required number of ways
                   =  6! / (1!)(2!)(1!)(1!)(1!)
                   = 6 * 5 * 4 * 3 * 2 *1 / 2 * 1
                   =360
                                                
9.How many arrangements can be made out of the letters of the word 
'MATHEMATICS' be arranged  so that the vowels always come
together

Sol:-           In the word '  MATHEMATICS' we treat vowels
           AEAI as one letter thus we have MTHMTCS(AEAI)
        now we have to arrange 8 letters out of which M occurs
           twice ,T occurs twice & the rest are different
          Number of ways of arranging these letters
                                = 8! / (2!)(2!)
                                = 10080
now AEAI has 4 letters in which A occurs 2 times and the rest 
are different 
          Number of ways of arranging these letters
                                            = 4! / 2! = 12
          Required number of words = (10080 * 12)
                                     = 120960

10.In how many different ways can the letter of the word 'DETAIL' be
arranged in such a way that the vowels occupy only the odd positions

Sol:-    These are 6 letters in the given word , out of which
             there are 3 vowels and 3 consonants
          Let us mark these positions as under

                           (1)(2) (3) (4)(5)(6)
  now 3 vowels can be placed at any of the three places out of 4
               marked 1,3,5
     Number of ways of arranging the vowels = 3P3 = 3! =6
  Also, the 3 consonants can be arranged at the remaining 3 positions
     Number of  arrangements  = 3P3 = 6
      Total number of ways = (6 * 6) =36

11.How many 3 digit numbers can be formed from the digits 2,3,5,6,7
 and 9 which are divisible by 5 and none of the digits is repeated?

Sol:- Since each desired number is divisible by 5,
         so we much have 5 at the unit place.   The hundreds place 
    can now be filled by any of the remaining 4 digits .so, there
         4 ways of filling it.
         Required number of numbers  = (1 * 5 * 4)
                                     = 20
12.In how many ways can 21 books on English and 19 books on Hindi
 be placed in a row on a self so that two books on Hindi may not 
 be together?

Sol:-        In order that two books on Hindi are never together,
        we must place all these books as under:
                          X E X E X . . . . . . . . . . X E X
        Where  E denotes the position of an English and X that of   
         a Hindi book.
        Since there are 21 books on English,the number of places
        marked X are therefore 22.
        Now, 19 places out of 22 can be chosen in
                      22 C 19 = 22 C 3 =22 * 21 * 20 / 3 * 2 *1
        Hence the required number of ways = 1540

13.Out of 7 constants and 4 vowels how many words of 3 consonants
  and 2 vowels can be formed?

Sol:-     Number of ways of selecting (3 consonants out of 7) and 
                                        (2 vowels out of 4)
                        = 7C3 * 4C2
                         = 210
   Number of groups each having 3 consonants and 2 vowels = 210
               Each group contains 5 letters
      Number of ways of arranging 5 letters among themselves
                        = 5! = (5 * 4 * 3 * 2 * 1)
                               = 210
           Required number of words = (210 * 210)
                                    = 25200