GMAT / GRE Mathematics Tutorial Series: Permutations And Combinations
In this article, we’ll cover the basic definitions, formulas and solved examples of Permutation and combination.
Definition: Permutation is the process of arranging the members of a set into a sequence. For e.g.
The different ways in which the alphabets L, M and N can be grouped together, taken all at a time, are LMN, LNM, MNL, MLN, NLM, and NML.
Note that LMN and NML are not the same as the order of arrangement is different. This condition is applicable for solving any problem in Permutations.
The number of ways in which n things can be arranged, taken all at a time, is given by
nPn = n!, also called ‘n factorial.’
Factorial of a number n is defined as the product of all the numbers from n to 1.
For example, the factorial of 7, 7! = 7*6*5*4*3*2*1 = 5040.
Therefore, the number of ways in which the 5 letters can be arranged, taken all a time, is 5! = 5*4*3*2*1 = 120 ways.
The number of permutations of n things, taken r at a time, denoted by:
nPr = n! / (n-r)!
The different ways in which the 4 letters, taken 2 at a time, can be arranged is 4!/(4-2)! = 4!/2! = 12 ways.
Important Permutation Formulae
1! = 1
0! = 1
Solved Permutation Examples
Problem 1: Find the number of words, with or without meaning that can be formed with the letters of the word ‘GLOBE.
‘GLOBE contains 5 letters.
Therefore, the number of words that can be formed with these 5 letters = 5! = 5*4*3*2*1 = 120 words.
Problem 2: Find the number of words, with or without meaning that can be formed with the letters of the word ‘EVENT’.
The word ‘EVENT’ contains 5 letters and the letter ‘E’ comes twice.
When a letter repeats itself more than once in a word, the factorial of the number of all letters in the word is divided by the number of occurrences of each letter.
Therefore, the number of words formed by ‘EVENT’ = 5!/2! = 60.
Problem 3: Find the number of words, with or without meaning, which can be formed with the letters of the word ‘FOOTBALL?
The word ‘FOOTBALL’ contains 8 letters. Of which, O occurs twice and L occurs twice.
Therefore, the total number of words formed = 8! / (2!*2!) = 10080 words.
Problem 4: How many different words can be formed with the letters of the word ‘ANGEL’ such that the vowels always come together?
The word ‘ANGEL’ contains 5 letters.
We have to find the number of permutations that can be formed where the two vowels A and E come together.
In cases like these, we group the letters that should come together and consider that group as a single letter.
So, the letters are N, G, L (A, E). Now the number of words is 4.
Therefore, the number of ways in which 4 letters can be arranged is 4!
For A and E, the number of ways in which A and E can be arranged is 2!
Hence, the total number of ways in which the letters of the ‘ANGEL’ can be arranged such that vowels are always together are 4! * 2! = 48 ways.
Problem 5: Find the number of different words that can be formed with the letters of the word ‘CUTTER’ so that the vowels are always together.
The word ‘CUTTER’ contains 6 letters.
The letters U and E should always come together. So the letters are C, T, T, R, (UE).
The number of ways in which the letters above can be arranged = 5!/2! = 60 (since the letter ‘T’ is repeated twice).
Number of ways in which U and E can be arranged = 2! = 2 ways
Therefore, total number of permutations possible = 60*2 = 120 ways.
Problem 6: Find the number of permutations of the letters of the word ‘SURNAME’ such that the vowels always occur in odd places.
The word SURNAME has 7 letters.
There are 4 consonants and 3 vowels in it.
Writing in the following way makes it easier to arrive at the answer.
(1) (2) (3) (4) (5) (6) (7)
No. of ways 3 vowels can occur in 4 different places = 4P3 = 24 ways.
After 3 vowels take 3 places, no. of ways 4 consonants can take 4 places = 4P4 = 4! = 24 ways.
Therefore, total number of permutations possible = 24*24 = 576 ways.
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A combination determines the number of possible arrangements from a collection in any order.
The different selections possible from the alphabets Q, R, S, taken 2 at a time, are QR, RS and SQ.
It does not matter whether we select Q after R or R after Q. The order of selection is not important in combinations.
To find the number of combinations possible from a given group of items n, taken r at a time, the formula, denoted by nCr is
nCr = n! / [r! * (n-r)!]
As per the above example, the different selections possible from the alphabets Q, R, S, taken two at a time are
3C2 = 3! / (2! * (3-2)!) = 3 possible selections (i.e., QR, RS and SQ)
Important Combination Formulae
nCn = 1
nC0 = 1
nC1 = n
nCr = nC(n-r)
The number of selections possible with X, Y, Z, taken all at a time is 3C3 = 1 (i.e. XYZ)
Solved Combination Examples
Here are a few examples to understand combination.
Problem 1: In how many ways can a team of 1 boy and 3 girls be formed from a group of 3 boys and 4 girls?
No. of ways 1 boy can be selected from a group of 3 boys = 3C1 = 3! / 1!*(3-1)! = 3 ways.
No. of ways 3 girls can be selected from a group of 4 girls = 4C3 = 4! / (3!*1!) = 4 ways.
Problem 2: Among a set of 5 black socks and 3 red socks, how many selections of 5 socks can be made such that at least 3 of them are black socks.
Selecting at least 3 black socks from a set of 5 black socks in a total selection of 5 socks can be
3 B and 2 R
4 B and 1 R and
5 B and 0 R socks.
Therefore, our solution expression looks like this.
5C3 * 3C2 + 5C4 * 3C1 + 5C5 * 3C0 = 46 ways.
Problem 3: How many 4 digit numbers that are divisible by 10 can be formed from the numbers 1, 5, 4, 8, 9, 0 such that no number repeats?
If a number is divisible by 10, its units place should contain a 0.
_ _ _ 0
After 0 is placed in the units place, the tens place can be filled with any of the other 5 digits.
Selecting one digit out of 5 digits can be done in 5C1 = 5 ways.
After filling the tens place, we are left with 4 digits. Selecting 1 digit out of 4 digits can be done in 4C1 = 4 ways.
After filling the hundreds place, the thousands place can be filled in 3C1 = 3 ways.
Therefore, the total combinations possible = 5*4*3 = 60.
1) Solve the following.
- i) 30P2
A. 870, 435
B. 435, 870
C. 870, 470
D. 435, 835
30P2 = 30! / 28! = 30*29*28! / 28! = 30*29 = 870.
30C2 = 30! / (2!*28!) = 435.
2) How many different possible permutations can be made from the word ‘MALLET’ such that the vowels are never together?
The word ‘MALLET’ contains 6 letters of which 1 letter occurs twice = 6! / 2! = 360
No. of permutations possible with vowels always together = 5! * 2! / 2! = 120
No. of permutations possible with vowels never together = 360-120 = 240.
3) In how many ways can a team of 3 boys and 2 girls can be made from a group of 5 boys and 5 girls?
Explanation: 5C3 * 5C2 = 100
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