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
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 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)
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
