PERMUTATIONS AND COMBINATIONS

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PERMUTATIONS AND COMBINATIONS

IMPORTANT FACTS AND FORMULAE


Factorial Notation: Let n be a positive integer.  Then, factorial n, denoted by n!  is defined as:
                           
                        n! = n(n-1)(n-2)........3.2.1.

Examples: (i) 5! = (5x 4 x 3 x 2 x 1) = 120; (ii) 4! = (4x3x2x1) = 24 etc.
We define, 0! = 1.

Permutations:  The different arrangements of a given number of things by taking some or all at a time, are called permutations.

Ex. 1.All permutations (or arrangements) made with the letters a, b, c by taking  two at a time are: (ab, ba, ac, bc, cb).

Ex. 2.All permutations made with the letters a,b,c, taking all at a time are:
(abc, acb, bca, cab, cba).

Number of Permutations: Number of all permutations of n things, taken r  at a time, given by:

                        nPr  = n(n-1)(n-2).....(n-r+1) = n!/(n-r)!

Examples: (i) 6p2 = (6x5) = 30. (ii) 7p3 = (7x6x5) = 210.

Cor. Number of all permutations of n things, taken all at a time = n!

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 the different groups or selections which can be formed by taking some or all of a number of objects, is called a combination.

Ex. 1. Suppose we want to select two out of three boys A, B, C.  Then, possible selections are AB, BC and CA.
Note that AB and BA represent the same selection.

Ex. 2. All the combinations formed by a, b, c, taking two at a time are ab, bc, ca.

Ex. 3. The only combination that can be formed of three letters a, b, c taken all at a time is abc.

Ex. 4. Various groups of 2 out of four presons A, B, C, D are:

                        AB, AC, AD, BC, BD, CD.

Ex. 5. Note that ab and ba are two different permutations but they represent  the same combination.

Number of Combinations: The number of all combination of n things,
taken r at a time is:

            nCr = n! / (r!)(n-r)! = n(n-1)(n-2).....to r factors / r!

Note that: ncr ­ = 1 and nc0 = 1.

An Important Result:  ncr = nc(n-r).

Example:  (i)  11c4 = (11x10x9x8)/(4x3x2x1) = 330.

                 (ii) 16c13 = 16c(16-13) = 16x15x14/3! = 16x15x14/3x2x1 = 560.

SOLVED EXAMPLES


Ex. 1. Evaluate: 30!/28!

Sol.  We have, 30!/28! = 30x29x(28!)/28! = (30x29) = 870.

Ex. 2. Find the value of (i) 60p3  (ii) 4p4

Sol.  (i) 60p3 = 60!/(60-3)! = 60!/57! = 60x59x58x(57!)/57! = (60x59x58) = 205320.


         (ii) 4p4 = 4! = (4x3x2x1) = 24.

Ex. 3. Find the vale of (i) 10c3 (ii) 100c98 (iii) 50c50

Sol. (i) 10c3 = 10x9x8/3! = 120.

        (ii) 100c98 = 100c(100-98) = 100x99/2! = 4950.

        (iii) 50c50 = 1.      [ncn = 1]

Ex. 4. How many words can be formed by using all letters of the word “BIHAR”

Sol.  The word BIHAR contains 5 different letters.

Required number of words = 5p5 = 5! = (5x4x3x2x1) = 120.

Ex. 5. How many words can be formed by using all letters of the word ‘DAUGHTER’ so that the vowels always come together?

Sol.  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 letters to be arranged are DGNTR (AUE).

Then 6 letters to be arranged in 6p6 = 6! = 720 ways.

The vowels in the group (AUE) may be arranged in 3! = 6 ways.

Required number of words = (720x6) = 4320.

Ex. 6. How many words can be formed from the letters of the word ‘EXTRA’ so that the vowels are never together?

Sol.  The given word contains 5 different letters.
        Taking the vowels EA together, we treat them as one letter.
        Then, the letters to be arranged are XTR (EA).
        These letters can be arranged in 4! = 24 ways.
        The vowels EA may be arranged amongst themselves in 2! = 2 ways.
        Number of words, each having vowels together = (24x2) = 48 ways.
       Total number of words formed by using all the letters of the given words
                           = 5! = (5x4x3x2x1) = 120.

Number of words, each having vowels never together = (120-48) = 72.


Ex. 7.  How many words can be formed from the letters of the word ‘DIRECTOR’
So that the vowels are always together?

Sol. In the given word, we treat the vowels IEO as one letter.
       Thus, we have DRCTR (IEO).
       This group has 6 letters of which R occurs 2 times and others are different.
       Number of ways of arranging these letters = 6!/2! = 360.
       Now 3 vowels can be arranged among themselves in 3! = 6 ways.
       Required number of ways = (360x6) = 2160.

Ex. 8.  In how many ways can a cricket eleven be chosen out of a batch of
15 players ?

Sol.  Required number of ways = 15c11 = 15c(15-11) = 11c4

                                                   = 15x14x13x12/4x3x2x1 = 1365.

Ex. 9.  In how many ways, a committee of 5 members can be selected from
6 men and 5 ladies, consisting of 3 men and 2 ladies?

Sol.  (3 men out 6) and (2 ladies out of 5) are to be chosen.

Required number of ways = (6c3x5c2) = [6x5x4/3x2x1] x [5x4/2x1] = 200.    



1. 
From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?
A.
564
B.
645
C.
735
D.
756
E.
None of these
2. 
In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together?
A.
360
B.
480
C.
720
D.
5040
E.
None of these
3. 
In how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together?
A.
810
B.
1440
C.
2880
D.
50400
E.
5760
4. 
Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?
A.
210
B.
1050
C.
25200
D.
21400
E.
None of these
5. 
In how many ways can the letters of the word 'LEADER' be arranged?
A.
72
B.
144
C.
360
D.
720
E.
None of these

In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?
A.
159
B.
194
C.
205
D.
209
E.
None of these
7. 
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?
A.
5
B.
10
C.
15
D.
20
8. 
In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women?
A.
266
B.
5040
C.
11760
D.
86400
E.
None of these
9. 
A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw?
A.
32
B.
48
C.
64
D.
96
E.
None of these
10. 
In how many different ways can the letters of the word 'DETAIL' be arranged in such a way that the vowels occupy only the odd positions?
A.
32
B.
48
C.
36
D.
60
E.
120
11. 
In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?
A.
63
B.
90
C.
126
D.
45
E.
135
12. 
How many 4-letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed?
A.
40
B.
400
C.
5040
D.
2520
13. 
In how many different ways can the letters of the word 'MATHEMATICS' be arranged so that the vowels always come together?
A.
10080
B.
4989600
C.
120960
D.
None of these
14. 
In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together?
A.
120
B.
720
C.
4320
D.
2160
None of these


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