30. 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.
