All about Permutation and Combination Permutation
Permutations
are different numbers of arrangements. Making words of 4 letters
without repetition or with repetition is an example of permutation. Now
suppose we have to make a word of 4 letters by taking any 4 letters from
10 letters A, B, C, D, E, F, G, H, I
☞ For example, permutations of 4 distinct objects out of 20 distinct objects can be written as
Combination
A
combination is a grouping or selection of all or part of a number of
things without reference to arrangement of the things selected.
For
example selection of three members from a family of ten, number of
cricket matches in a group having 10 teams, number of triangles formed
by joining any three points out of 20 points in plane etc.
☞ The symbol
Suppose
there are 5 students and a company wants to select any 3 of them. So
the total number of ways = selection of 3 out of 5 = rejection of 2 out
of 5.
Word Formation
Word formation is based on fundamental principle of multiplication.
☞ For
example, total number of words which can be formed by using the letters
of the word “ACHILLES” will be 8!/2!, as the letter L appeared twice.
Gap Method
Gap
method is used if we have to arrange ‘n’ different things in such a way
such that no two of the ‘r’ things are together. In this method we first
place (n – r) things in a line so that there are (n – r + 1) gaps in
between and at extreme ends. Now place ‘r’ things on these gaps so that
no two of these things are together provided that r ≤ n – r + 1. Suppose
T1,T2,T3,………… T10 are 10 distinct objects and we want that no two of
T1,T2,T3 and T4 are together in the arrangement. First place remaining
things T5,T6……… T10
_T5_T6_T7_T8_T9_T10_
So there are 7 gaps
Total number of ways = (7.6.5.4).6!
Rank of a Word in Dictionary
If all
the words which can be made by using n different letters are arranged
in a dictionary in alphabetical order then position at which a
particular word appears, is known as its rank. For example 24 words can
be made by using letters of the word “RANK” without repetition. Now to
calculate position of the word “RANK” we arrange the words in
alphabetical order.
Hence position of the word RANK is:
3! + 3! + 3! + 1 + 1 = 20
Sum of all the numbers which can be formed by using the digits without repetition
Suppose the digits are 1, 2, 3, 4. We know that total number of numbers which can be formed without repetition is 4! = 24. Now
Numbers with unit digit 1 will be 3! = 6
Numbers with unit digit 2 will be 3! = 6
Numbers with unit digit 3 will be 3! = 6
Numbers with unit digit 4 will be 3! = 6
Hence sum of the unit digits
= 6 × (1 + 2 + 3 + 4) = 60,
Similarly sum of the digits at tens, hundred’s and thousand’s place = 60
Sum of all the numbers
= 60 × 1 + 60 × 10 + 60 × 100 + 60 × 1000 = 66660
Sum of
all the numbers which can be formed by using the ‘n’ non zero digits
without repetition is (n – 1)! (sum of the digits) (111……n times).
Circular Permutation
☞Number of circular permutations of n distinct objects = (n – 1)!
☞Number of circular permutations of n distinct objects =
if there is no difference between clock wise and anticlockwise permutations.
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