## Permutations and Combinations: Quantitative Aptitude Notes

PERMUTATION AND COMBINATION
Permutation: The different arrangements of a given number or things by taking some or all at a time, are called Permutations.
a) All permutations (or arrangements) made with the letters of a, b, c by taking two at a time are (ab, ba, ac, ca, bc, cb)
b) 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:
npr = n ( n – 1 ) ( n – 2 )……(n – r + 1 ) = n! / ( n – r ) !
1. Ex.6p2 = (6 x 5) = 30
2. Ex.7p3 = (7 x 6 x 5) = 210;
Note: Number of all permutation that is n things, we can take all at a time = n!
An Important Result:
If there are n subjects 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!) (p3!)……………. (pr!)]
Example: In how many different ways the words ‘HOUSE’ can be arranged?
Answer: The word HOUSE contains 5 letters.
Required number of word using formula 5p5 = 5! = (1 x 2 x 3 x 4 x 5) = 120.
Example: In how many several way the word KOLKATA can be prepared so that as vowel always come together?
First of all we need to know how many vowel in this given word, here is three vowel that is O and A, A. then we count the consonant that is four. Now we count the number of vowel as a single unit that is vowel O A, A count as single unit and add it with consonant so we have a four unit.
(4 consonant +three vowels as single unit) = 5 x 3(three vowel)
5! x 3! = 1 x 2 x 3 x 4 x 5 x 1 x 2 x 3 = 720 ways we prepared the word.
Example: In how many several way the word HOUSE can be prepared so that as vowel always come together?
First of all we need to know how many vowel in this given word, here is three vowel that is E ,O and U. Then we count the consonant that is Two. Now we count the number of vowel as a single unit that is vowel E,O and U count as single unit and add it with consonant so we have a three unit.
(2 consonant + three vowels as single unit) = 3 x 3(three vowel)
3! x 3! = 1 x 2 x 3 x 1 x 2 x 3 = 36 ways we prepared the word.
Example: In how many several ways the word CAME can be arranged, that the vowels not come together?
First of all we need to know how many vowel in this given word, here is two vowel that is A and E. Then we count the consonant that is Two. Now we count the number of vowel as a single unit that is vowel A and E count as single unit and add it with consonant so we have a three unit.
(2 consonant + two vowels as single unit) = 3 x 2(two vowel)
Note: If we Subtract total number ways from within vowel ways than we get the not come to-gather ways,
So, (4! – 3! x 2! ) = 12.
What is Combination?
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. In combination we can select only one item at time it based on selection on choice. Suppose Ball, Marbel, etc.
Examples:
Type 1: Suppose we want to select two out of three girls A, B, C. Then possible selections are AB, BC, and CA.
Note: that AB and BA represent the same selection.
Type 2: Suppose we select all three girls Like A, B, C. Then the selection is ABC.
Note: BAC, ABC, CAB are the same choice of selection.
Type 3: All the combinations formed by a, b, c, taking two at a time are ab, bc, ca.
Type4: The only combinations that can be formed of three letters a, b, c taken all at a time is abc.
Type5: Various persons of 2 out of four persons A, B, C and D are: AB, AC, AD, BC, BD, CD.
Type6: Note that ab and ba are two different permutations but they represent the same combination.
Number of Combinations: The number of all combinations of n things, taken r at a time is:
nCr = n! / r! ( n – r ) ! = n ( n – 1 ) ( n – 2 ) …….to r factors / r!.
Note: If n = r then nCr = 0 and nC0 = 1
Important Rule:
nCr = nC(n – r ) = 1
Example: find the value of 16C13
16C( 16 – 13 ) = 16C3 = (16 x 15 x 14 x 13!)/ (3!)
= (16 x 15 x 14) / (3 x 2 x 1) = 560
Example: Find out value of 10P3
10C3= 10 x 9 x 8 / 3! = 10 x 9 x 8 / 3 x 2 x 1 = 120
Example: Find the value of 50C50
50C50 = 1. [nCn = 1]
Difference between Permutations and Combinations and How to identify them
Sometimes, it will be clearly stated in the problem itself whether permutation or combination is to be used. However if it is not mentioned in the problem, we have to find out whether the question is related to permutation or combination.
Consider a situation where we need to find out the total number of possible samples of two objects which can be taken from three objects P, Q, R. To understand if the question is related to permutation or combination, we need to find out if the order is important or not.
a) If order is important, PQ will be different from QP, PR will be different from RP and QR will be different from RQ
b) If order is not important, PQ will be same as QP, PR will be same as RP and QR will be same as RQ
Hence,
a) If the order is important, problem will be related to permutations.
b) If the order is not important, problem will be related to combinations.
For permutations, the problems can be like "What is the number of permutations the can be made", "What is the number of arrangements that can be made", "What are the different number of ways in which something can be arranged", etc.
For combinations, the problems can be like "What is the number of combinations the can be made", "What is the number of selections the can be made", "What are the different number of ways in which something can be selected", etc.
pq and qp are two different permutations, but they represent the same combination.
Mostly problems related to word formation, number formation etc will be related to permutations. Similarly most problems related to selection of persons, formation of geometrical figures, distribution of items (there are exceptions for this) etc will be related to combinations.
Repetition:
The term repetition is very important in permutations and combinations. Consider the same situation described above where we need to find out the total number of possible samples of two objects which can be taken from three objects P, Q, R.
a) If repetition is allowed, the same object can be taken more than once to make a sample. i.e., PP, QQ, RR can also be considered as possible samples.
b) If repetition is not allowed, then PP, QQ, RR cannot be considered as possible samples. Normally repetition is not allowed unless mentioned specifically.

MEMORY BASED SOLVED EXAMPLES BASED ON VARIOUS TYPES:
Type 1 - Based on Selection
1. A boy has 3 library cards and 8 books of his interest in the library. Of these 8, he does not want to borrow chemistry part II unless Chemistry part I is also borrowed. In how many ways can he choose the three books to be borrowed?
Solution:
Two possibilities are there:
(i) Chemistry part I is available in 8 books with Chemistry part II
Or
(ii) Chemistry part II is not available in 8 books but Chemistry part I is available.
Total No. of was = 1 x 1 x 6C1 + 7C3 = 6 + (7 x 6 x 5) / (3 x 2) = 6 + 35 = 41
2. In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?
Solution:
In a group of 6 boys and 4 girls, four children are to be selected such that at least one boy should be there.
Hence we have 4 options as given below
We can select 4 boys ...(option 1)
Number of ways to this = 6C4
We can select 3 boys and 1 girl ...(option 2)
Number of ways to this = 6C3 × 4C1
We can select 2 boys and 2 girls ... (option 3)
Number of ways to this = 6C2 × 4C2
We can select 1 boy and 3 girls ... (option 4)
Number of ways to this = 6C1 × 4C3
Therefore total number of ways = 6C4 + 6C3 × 4C1 + 6C2 × 4C2 + 6C1 × 4C3 = 15 + 18 + 90 + 24 = 209
3. A box contains 1010 balls out of which 33 are red and rest are blue. In how many ways can a random sample of 66 balls be drawn from the bag so that at the most 22 red balls are included in the sample and no sample has all the 66 balls of the same colour?
Solution:
Six balls can be selected in the following ways:
One red ball and 5 blue ball (Or) Two red balls and 4 blue balls
Total number of ways = (3C1 × 7C6) + (3C2 × 7C4) = 63 + 105 = 168
4. How many groups of 6 persons can be formed from 8 men and 7 women?
Solution:
Total no. of person = 8 + 7 = 15
No. of groups = 15C6 = 15! /{6! (15 - 6)!} = 15! / (6! 9!) = (15 x 14 x 13 x 12 x 11 x 10) × 9! /(9! ×6 x 5 x 4 x 3 x 2 x 1) = 5005
5. There is a 7-digit telephone number with all different digits. If the digit at extreme right and extreme left are 5 and 6 respectively, find how many such telephone number are possible?
Solution:
There is a 7 - digit telephone number but extreme right and extreme left positions are fixed.
i.e. for first place and last we have choice respectively i.e., 6 x x x x x 5
For 2nd place we have 8 choices (i.e. 0,1,2,3,4,7,8,9,)
similarly for next place we have 7 choices ,
for next place we have 6 choices and so on...
Required number of ways = 8C1 x 7C1 x 6C1 x 5C1 x 4C1= 8 x 7 x 6 x 5 x 4 = 6720
6. In how many ways, can 24 persons be seated around a circular table if there are is 13 seats?
Solution:
First , we select 13 persons out of 24 persons in 24C13 ways.
Now, these 13 persons can be seated in 12! ways around a table .
So required number of ways = 24C13 x 12! = [24! / {13!(24 - 13)!}] x 12! = [24! / {13! x 11!}] x 12! = 24! × 12! / {13 × 12! × 11!} = 24! / (13 x 11!)
7. The number of ways in which a committee of 3 ladies and 4 gentlemen can be appointed from a group consisting of 8 ladies and 7 gentlemen, if Mrs. X refuses to serve in a committee if Mr. Y is its member, is
Solution:
If Mrs. X is selected among the ladies in the committee, then Mr. Y is not selected or if Mrs. X is not selected then Mr. Y can be there in the committee...
So, required number of ways = 8C3 x 6C4 + 7C3 x 7C4 = [(8 x 7 x 6)/(3 x 2)] x [(6 x 5)/(2 x 1)] + [(7 x 6 x 5)/(3 x 2)] x [(7 x 6 x 5)/(3 x 2)] = 840 + 1225 = 2065
Type 2 – Based on Selection + Arrangement
1. There are 5 yellow, 4 green and 3 black balls in a bag. All the 12 balls are drawn one by one and arranged in a row. Find out the number of different arrangements possible.
Solution:
Number of different arrangements possible = 12!/(5! × 4! × 3!) = [12 ×11 ×10 ×9 ×8 ×7 ×6 ×5 ×4 ×3 ×1/(5 ×4 ×3 4 ×3 ×2 3 ×2)]
= 11 ×10 ×9 ×4 ×7 = 27720
2. In how many ways can 9 different colour balls be arranged in a row so that black, white, red and green balls are never together?
Solution:
Total number of ways in which 9 different colour balls can be arranged in a row = 9! …….(A)
Now we will find out total number of ways in which 9 different colour balls can be arranged in a row so that black, white, red and green balls are always together.
We have total 9 balls. Since black, white, red and green balls are always together, group these 4 balls together and consider as a single ball. Hence we can take total number of balls as 6. These 6 balls can be arranged in 6! ways.
We had grouped 4 balls together. These 4 balls can be arranged among themselves in 4! ways.
Hence, total number of ways in which 9 different colour balls be arranged in a row so that black, white, red and green balls are always together = 6! × 4! …… (B)
From (A) and (B),
Total number of ways in which 9 different colour balls can be arranged in a row so that black, white, red and green balls are never together = 9! – 6! X 4! = 345600
3. A mixed doubles tennis game is to be played between two teams (each team consists of one male and one female). There are four married couples. No team is to consist of a husband and his wife. What is the maximum number of games that can be played?
Solution:
Married couples:
MF MF MF MF AB, CD, EF, GH
Possible teams:
AF CF ED GD
AH CH EH GF
Team AD can play only with CB, CF, CH, EB, EH, GB, GF (7 teams).
Teams AD cannot play with AF, AH, ED and GD.
The same will apply with all teams, So, number of total matches = 12 x 7 = 84
But every match includes 2 teams, so the actual number of matches = 84/2 = 48
4. In how many different ways can four books A, B, C and D be arranged one above another in a vertical order such that the books A and B are never in continuous position?
Solution:
The number of arrangement in which A and B are not together = Total number of arrangement - Number of arrangement in which A and B are together = 4! - 3! x 2! = 24 - 12 = 12
5. 2 Men and 1 women board a bus in which 5 seats are vacant, one of these 5 seats is reserved for ladies. A woman may or may not sit on the seat reserved for ladies. In how many different ways can the five seats be occupied by these passengers?
Solution:
Case I :-
If lady sets on reserved seat, then
2 men can occupy seats from 4 vacant seats in 4P2 = 4 x 3 = 12 ways
Case II :-
If lady does not site on reserved seat, then 1 women can occupy a seat from seat in 4 ways, 1 man can occupy a seat from 3 seats in 3 ways, also 1 man left can occupy a seat from remaining two seats in 2 ways.
Total ways = 4 x 3 x 2 = 24 ways
Hence, from Case I and case II, total ways = 12 + 24 = 36 ways
Type 3 – Word Problems
1. In how many different ways can the letters of the word ‘MATHEMATICS’ be arranged so that the vowels always come together?
Solution:
In MATHEMATICS, the consonants M and T are repeated two times each.
Also the vowel A is repeated two times.
Since there are four vowels being repeated, vowels can be arranged in 4!/2 = 12 ways
Now remaining 7 consonants, with M, T being repeated, can be written in 7!/(2 * 2) = 1260 ways
Now four vowels together can take any of the 8 places as shown below:
Total number of ways in which the letters of the word MATHEMATICS can be arranged such that vowels always come together = 1260 x 8 x 12 = 120960
2. In how many ways can the letters of the word 'Director' be arranged so that the three vowels are never together?
Solution:
Total number of letters = 8
Number of vowels = 3; r occurs twice.
Total number of arrangements when there is no restriction = 8! / 2!
When three vowels are together, regarding them as one letter, we have only 5 + 1 = 6 letters
These six letters can be arranged in 6! / 2! Ways, since r occurs twice.
But the three vowels can be arranged among themselves in 3! ways.
Hence number of arrangement when the three vowels are together = 6! x 3! / 2!
Required number = 8! / 2! - {6! x 3! / 2!} = 18,000
3. In how many different ways can the letters of word JUDGE be arranged so that the vowels always come together?
Solution:
Total number of letters = 5
Number of vowels = 2.
If we consider both vowel as a one letter then,
Required number = 4! 2! = 48.
4. How many three letter computer passwords can be formed (no repetition allowed) with at least one symmetric letter?
Solution:
Total number of password using all alphabets -Total
number of password using no symmetric alphabets = (26 x 25 x 24) - (15 x 14 x 13) = 12870
5. How many different numbers can be formed from the digits 3, 4, 5, 6 and 7 when repetitions are allowed?
Solution:
Number of 1 digit numbers = 5
Number of 2 digit numbers = 52 = 25
Number of 3 digit numbers = 53 = 125
Number of 4 digit numbers = 54 = 625
Number of 5 digit numbers = 55 = 3125
Total number of numbers formed with these digits
= 5 + 25 + 125 + 625 + 3125 = 3905
6. In how many ways, can the letters of the word 'ASSASSINATION' be arranged, so that all the S are together?
Solution:
When All S are taken together, then ASSASSINATION has 10 letters.
So, 10 letters in total can be arranged in 10 ! ways .
[ All 'S' are considered as 1]
But, here are 3 'A' and 2'I' and 2 'N' .
Required number of ways = 10! / (3! x 2! x 2!) = 151200
Type 4 – Miscellaneous
1. 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?
Solution:
Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.
The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.
The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.
Therefore, required number of numbers = (1 x 5 x 4) = 20.
3. There are 14 points in a plane, out of which 4 are collinear. Find the number of triangles made by these points.
Solution:
The required number of triangles = nC3 - mC3
Here, n = 14, m = 4
= 14C3 - 4C3
= (14 x 13 x 12 x 11! ) / (3! x 11!) - 4! / (3! x 1!)
= (14 x 13 x 12)/6 - 4/1
= 14 x 26 - 4 = 364 - 4 = 360
4. A question paper consists of two sections having respectively 3 and 5 questions. The following note is given on the paper. "It is not necessary to attempt the entire question". One question from each section is compulsory. In how many ways, a candidate can select the question?
Solution:
Here, we have two sections A and B. Section A has 3 question and B has 5 question and one question from each section is compulsory according to the given condition.
Number of ways selecting one or more than one question from section A = 23 - 1 = 7
Similarly, from section B = 25 - 1 = 31
According to the rule of multiplication, the required number of ways in which a candidate can select the question = 7 x 31 = 217
5. A five-digit number divisible by 3 is to be formed using digits 0, 1, 2, 3, 4 and 5 without repetition, the total number of ways this can be done, is
Solution:
A five-digit number, which is divisible by 3, is formed when sum of digits is also divisible by 3.
So, combination formed using six-digits, which are divisible by 3 = 5 + 4 + 3 + 2 + 1 = 15 = 5 + 4 + 2 + 1 + 0 = 12
So, set of number are (5, 4, 3, 2, 1) and (5, 4, 2, 1, 0).
Number formed by using 1st set = 5 x 4 x 3 x 2 x 1 = 120
Similarly, using 2nd set = 4 x 4 x 3 x 2 x 1 = 96
Hence, using 2nd set, underlined place cannot be filled by 0, otherwise it will become a four-digit number.
Total number = 120 + 96 = 216
6. A new flag is to be designed with six vertical stripes using some or all of the colour yellow, green,blue and red. Then, the number of ways this can be made such that no two adjacent stripes have the same colour is
Solution:
Any of the 4 colour can be chosen for the first stripe. Any of the remaining 3 colours can be used for the second stripe. The third stripe can again be coloured in 3 ways (we can repeat the colour of the first stripe but not use the colours of the second stripe).
Similarly, There are 3 ways to colour each of the remaining stripes.
The number of ways the flag can be coloured is 4(3)5 = (12) (3)4 = 12 x 81
7. How many numbers can be formed from 1, 2, 3, 4, 5, (without repetition), when the digit at the unit's place must be greater than that in the ten's place?
Solution:
The digit in the unit's place should be greater than that in the tens' place.
Hence, if digit 5 occupies the unit place, then remaining four digits need not to follow any order, hence required number = 4!
However, if digit 4 occupies the unit place then 5 cannot occupy the tenth position. Hence, digit at the ten's place and it will be filled by the digit 1, 2 or 3. This can happen in 3 ways. The remaining 3 digit can be filled in the remaining three place in 3! ways.
Hence, in all, we have (3 x 3!) numbers ending in 4. Similarly, if we have 3 in the unit's place and it will be either 1 or 2. This can happen in 2 ways. Hence, we will have (2 x 3!) number ending in 3 . Similarly, we can find that there will be 3! numbers ending in 2 and no number with 1. Hence, total number of numbers
= 4! + (3) x 3! + (2 x 3!) + 3! = 4! + 6 x 3! = 24 + (6 x 6) = 60
8. Everybody in a room shakes hands with everybody else. The total number of handshakes is 66. The total number of persons in the room is?
Solution:
Let there be n persons in the room. The total number of hand shakes is same as the number of ways of selecting 2 out of n.
nC2 = 66
n(n - 1) / 2! = 66
n2 - n - 132 = 0
(n - 12) (n + 11) = 0
n = 12