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Web Notes

Permutation

Quantitative Aptitude (HP Allied) Permutation & Combination

Permutation

Meaning of Permutation:- The notion of permutation relates to the act of arranging all the members of a set into some sequence or order, or if the set is already ordered, rearranging (reordering) its elements, a process called permutation.The different arrangements of a given number of things by taking some or all at a time, are called permutations.

Examples:

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

All permutations made with the letters a, b, c taking all at a time are:

( abc, acb, bac, bca, cab, cba)

Number of Permutations:

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

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

When a certain number of things always occur together
To know better, consider the following given example.
1) In how many ways can 7 people be seated in a row for a photograph, if two particular people always want to be together.

Solution:-Consider the two people who want to be together as 1 unit. The number of permutations of remaining 5 people + 1 unit (2 people) = six units is 6!
For each of these two people considered as one unit can be arranged in 2! Ways.
Therefore, Total permutations = 6! X 2!
= 720 X 2 = 1440 ways

When certain things of one type do not occur together
Example- In how many ways can 10 examination papers be arranged so that the best & worst papers never come together.
Solution:-Total ways in which best & worst are never together = Total arrangements – Two papers are always together
Total ways = 10!-(9! X 2!)
For (9! X 2!) Please refer Type 1 example.

Formation of numbers with digits (digits can be repeated)
Example- How many 3 digit numbers can be formed using digits 1, 2,5,6,8 so that the digits can be repeated.

Solution:-Three digit number consists of 1unit place, 1 tens place &1 hundred place. In the given problem there are 5 numabers, as it can be repeated any no.of times.
Each place can be filled by any one of 5 digits
Total numbers=5x5x5 =125
125 such 3 digits number can be formed.

Formation of numbers with digits (digits are not repeated)

Case1:- When there is no restriction

Example: - How many different 4 digit numbers can be formed using the digits 1,2,4,5,7,8,9 no digit being repeated in any number.

Solution:- The thousand place (Th) can be filled in 7 ways. For each of these the hundred, tenth, units place can be filled in 6, 5, 4 ways respectively
Total ways = 7x6x5x4=840 nos.

Case 2:-Numbers divisible by a particular number or even number or odd numbers
Example: - How many 4 digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 no digit being repeated.
Solution:- If the number is even it has to end with even number (2, 4 or 6 in this given question).Hence units place has 3 choices & thousandth , hundredth & tenth place can be filled in 5 ,4,3 ways respectively.
Therefore, total 4 digit even numbers = 5x4x3x3 =180

Word building (Alphabets not repeated)
Case 1:- When there is no restriction
Example: - Find the number of words that can be formed by using all the letters of the word “MONKEY”.

Solution:- The word MONKEY has 6 letters. The six different letters can be arranged in 6P6 = 6! = 720 ways.

Case 2:- When few letters/vowels/consonants occur together

Example: -In how many ways can the letters of word “LAUGHTER” be arranged so that the vowels are always together.

Solution:- The word LAUGHTER has 3 vowels & 5 consonants. Consider the 3 vowels as 1unit.Hence 1unit+ 5 consonants.
Total= 6 can be arranged in 6! Ways. For each of these, the 3 vowels can be arranged in 3! Ways.
Therefore, total words= 6! x3! =720x6=4320

Case 3:-When vowels/consonants occupy odd/even places

Example:- In how many ways can the letters of the word HEXAGON be arranged so that the vowels are always in even places.

Solution:- The word HEXAGON has 7 letters of which 3 are vowels. Since total letters is 7, there are 4 odd places & 3 even places.
The 3 vowels can be arranged in the 3 odd places in 3! Ways. For each of these the 4 consonants can be arranged in the 4 odd
Places in 4! Ways. Hence total ways = 3!x4!=6x24=144

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