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
Tuesday, 12 November 2013
Permutations and Combinations
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