Permutation And Combination

Permutation And Combination
Permutation And Combination refer to the arrangement of the object, tasks, numbers and groups etc. In various competitive exam of Bank, CAT, IIT and SSC etc and school examination of 11th and 12th Aptitude Questions are based on this topic. Some time it is the reason of the fear for students to crack examination. No need to worry as we are sharing Permutation and Combination Formula and Permutation and Combination Problems and Solutions. You can easily have good command on this topic by qualifying this exam.

Permutation And Combination Definition

What is Permutation?
It defines each of several possible ways in which a set or number of things can be ordered or arranged. When the order does matter it is a Permutation. There are 2 types of Permutation.
First one is in which repetition is allowed: It could be “9999”.
Second one in which these is no Repetition: Order of 16 pool balls
What is Combinations?
Combination is joining or merging of different things without thinking their order. For example there is no order in fruit salad that is combination of apples, grapes and bananas. Combinations are also of two types.
First one is in which repetition is allowed: It could be (3,3,3,3, 9,9)
Second one in which these is no Repetition: Order of (1,4,5,7,3,99) pool balls

Permutation And Combination Important Formula and Explanation

Factorial Notation
Let n be a positive integer. Then, factorial n, denoted n! is defined as:
n! = n(n – 1)(n – 2) … 3.2.1.
Examples:
We define 0! = 1.
6! = (6 x 5x 4 x 3 x 2 x 1) = 24.
3! = (3 x 2 x 1) = 120.

Permutations
Permutations is the different arrangement of the given number.
Examples:
All arrangements made with the three letters p, q, r by taking two at a time are (pq, qp, pr, rp, qr, rq).
All arrangements made with the three letters p, q, r taking all at a time are: (pqr, prq, qpr, qrp, rpq, rqp)
Number of Permutations
Total Number of all permutations of n things,
nPr = n(n – 1)(n – 2) … (n – r + 1) = n!/ (n – r)!
Examples:
8P3 = (8 x 7 x 6) = 30.
5P3 = (7 x 6 x 5) = 210.
Cor. number of all permutations of n things, taken all at a time = n!
Important Result Used in Problems
If there are n subjects of which q1 are alike of one kind; q2 are alike of another kind; q3 are alike of third kind and so on and pr are alike of qth kind,
such that (q1 + q2 + … qr) = n.
Then, number of permutations of these n objects is = n!
(q1!).(q2)!…..(qr!)
Combinations
Combination is joining of different groups without thinking their order.
Examples:
If we desire to select two out of three boys P, Q, R, Then possible selections are PQ, QR and RA.
Combination of 3 letters P, Q, R taken all at a time is PQR only
6.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!
Important Result for Problems
nCn = 1 and nC0 = 1
nCr = nC(n – r)
Examples:
11C4 = (11 x 10 x 9 x 8)/ (4 x 3 x 2 x 1) = 330.
16C13 = 16C(16 – 13) = 16C3 = (16 x 15 x 14) / 3! = (16 x 15 x 14)/( 3 x 2 x 1) = 560.
8.Permutation And Combination Shortcuts (Special Case)
0! = 1
1! = 1
The Multiplication Theorem
If you perform a task in x different ways while other task is performed in y different ways, then the two operations in succession can be performed in x × y different ways.
Example:
Ram has 4 balls and Rekha has 4 bats, then how many pairs can be made.
For the balls we have 4 choices and same for bats we have 4 choices. So total pairs are the multiplication of the choices that is 4×4 = 16
Addition Theorem:
If you perform a task in x different ways while other task is performed in y different ways, then the two operations in succession can be performed in (x + y) different ways
LET'S TAKE A SOME EXAMPLE

In how many different ways can the letters of the word ‘CORPORATION’ be arranged so that the vowels always come together?
810
1440
2880
50400
5760
Answer: D
Explanation:
In the word ‘CORPORATION’, we treat the vowels OOAIO as one letter.
Thus, we have CRPRTN (OOAIO).
This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.
Number of ways arranging these letters = 7!/ 2! = 2520.
Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged
In 5!/ 3! = 20 ways.
Required number of ways = (2520 x 20) = 50400.
In how many ways can the letters of the word ‘LEADER’ be arranged?
72
144
360
720
None of these
Answer: C
Explanation:
The word ‘LEADER’ contains 6 letters, namely 1L, 2E, 1A, 1D and 1R.
Required number of ways = 6! / (1!)(2!)(1!)(1!)(1!) = 360
In how many different ways can the letters of the word ‘OPTICAL’ be arranged so that the vowels always come together?
120
720
4320
2160
None of these
Answer: B
Explanation:
The word ‘OPTICAL’ contains 7 different letters.
When the vowels OIA are always together, they can be supposed to form one letter.
Then, we have to arrange the letters PTCL (OIA).
Now, 5 letters can be arranged in 5! = 120 ways.
The vowels (OIA) can be arranged among themselves in 3! = 6 ways.
Required number of ways = (120 x 6) = 720.
In how many different ways can the letters of the word ‘DETAIL’ be arranged in such a way that the vowels occupy only the odd positions?
32
48
36
60
120
Answer: C
Explanation:
There 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 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 ways of these arrangements = 3P3 = 3! = 6.
Total number of ways = (6 x 6) = 36.
In how many different ways can the letters of the word ‘LEADING’ be arranged in such a way that the vowels always come together?
360
480
720
5040
None of these
Answer: C
Explanation:
The word ‘LEADING’ has 7 different letters.
When the vowels EAI are always together, they can be supposed to form one letter.
Then, we have to arrange the letters LNDG (EAI).
Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways.
The vowels (EAI) can be arranged among themselves in 3! = 6 ways.
Required number of ways = (120 x 6) = 720.
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?
5
10
15
20
Answer: D
Must Check: 11 Tips To Remember What You Read
Explanation:
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.
Required number of numbers = (1 x 5 x 4) = 20.
In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?
63
90
126
45
135
Answer: A
Explanation:
Required number of ways = (7C5 x 3C2) = (7C2 x 3C1) = {(7 x 6)/(2 x 1) x 3} = 63
How many 4-letter words with or without meaning, can be formed out of the letters of the word, ‘LOGARITHMS’, if repetition of letters is not allowed?
40
400
5040
2520
Answer: C
Explanation:
‘LOGARITHMS’ contains 10 different letters.
Required number of words = Number of arrangements of 10 letters, taking 4 at a time.
= 10P4
= (10 x 9 x 8 x 7)
= 5040.
In how many different ways can the letters of the word ‘MATHEMATICS’ be arranged so that the vowels always come together?
10080
4989600
120960
None of these
Answer: C
Explanation:
In the word ‘MATHEMATICS’, we treat the 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 and 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 x 12) = 120960.
10: In how many ways can 6 students of a school be seated around a circular table?
Answer: The problem is a cyclic permutation
The number of ways the 6 students can be seated
= 1 × (6 – 1) ! = 5! = 5 × 4 × 3 × 2 × 1 = 120
11: How many different 3 letter words can be made by 5 vowels, if vowel ‘A’ will never be included?
Answer: Total letters (n) = 5
So total number of ways = n-1Pr = 5-1P3 = 4P3 = 24.
Number of permutations of n different things taking all at a time, in which m specified things always come together = m!(n-m+1).
Five persons entered a lift cabin on the ground floor of an 8-floor house. Suppose that each of them can leave the cabin independently at any floor beginning with the first. What is the total number of ways in which each of the five persons can leave the cabin at any of the 7 floors? Answer: Any one of the 5 persons can leave the cabin in 7 ways independent of other.
Hence the required number of ways = 7 × 7 × 7 × 7 × 7 = 7⁵.
In how many ways 8 persons can be arranged in a circle?
Answer: The eight persons can be arranged in a circle in
(8 – 1) ! = 7! = 5040.
14: In how many ways can 5 prizes be given away to 4 boys, when each boy is eligible for all the prizes?
Answer: Any one of the prizes can be given in 4 ways; then any one of the remaining 4 prizes can be given again in 4 ways, since it may even be obtained by the boy who has already received a prize.
Hence 5 prizes can be given 4 × 4 × 4 × 4 × 4 = 4⁵ ways.
15. In how many ways a hockey team of eleven can be elected from 16 players?
Answer: Total number of ways
= 16C11 = 16!/ (11! x 5!)= 4368
= (16 x 15 x 14 x 13 x 12)/ (5 x 4 x 3 x 2) = 4368