The number of arrangements of the letters a b c d in which neither a, b nor c, d come together.
A) 6
B) 12
C) 16
D) None of these
Answer
Hint: In this question first we find the total number of ways to arrange 4 letters after that we find the number of possible ways in which \[{\text{a & b}}\] and \[{\text{c & d}}\] together. After that, we subtract it from the total number of ways and we will get the required answer.
Complete step by step solution: As we all know letters a b c d can be arranged in \[4!\]
\[4! = 4 \times 3 \times 2 \times 1\]
\[ \Rightarrow 4! = 24\]
\[ \Rightarrow \] Total
possible arrangement of letters a b c d is 24.
In order to calculate neither \[{\text{a & b}}\] nor \[{\text{c & d}}\] come together first we calculate the cases when \[{\text{a & b}}\] comes together and \[{\text{c & d}}\] together.
So, take \[{\text{a & b}}\] as one entity so we have total 3 letters.
\[ \Rightarrow \] 3 letters can be arranged in \[3!\] and \[{\text{a & b}}\] itself arranged in \[2!\]
\[ \Rightarrow \] Case when \[{\text{a & b}}\] comes
together then letters can be arranged in \[3! \times 2!\]\[ = 3 \times 2 \times 1 \times 2 \times 1 = 12\] ways.
Similarly,
Take \[{\text{c & d}}\] as one entity so we have total 3 letters.
\[ \Rightarrow \] 3 letters can be arranged in \[3!\] and \[{\text{c & d}}\] itself arranged in \[2!\]
\[ \Rightarrow \] Case when \[{\text{c & d}}\] comes together then letters can be arranged in \[3! \times 2!\]\[ = 3 \times 2 \times 1 \times 2 \times 1 = 12\] ways.
Now, the case
when \[{\text{a & b}}\] and \[{\text{c & d}}\] both come together.
\[ \Rightarrow \] Case when \[{\text{a & b}}\] and \[{\text{c & d}}\] both come together then letters can be arranged in \[2! \times 2! \times 2! = 2 \times 2 \times 2 = 8\] ways.
So, the total number of arrangement when neither \[{\text{a & b}}\] nor \[{\text{c & d}}\] come together \[ = 24 - 12 - 12 + 8\]
\[ = 8\]
8 number of ways the letters a b c d can arrange in which neither a,b nor c,d
come together.
Hence, option D. None of these is the correct answer.
Note: Here we have used the concept of permutation. Permutation can be defined as arranging or rearranging the elements of a set. It gives the number of ways a certain number of objects can be arranged. Combinations can be defined as selecting a number of elements from a given set. Both permutations and combinations are similar except that in permutations, the order is important and in combinations the order is not important.
In this article you’ll learn about Permutation and Combination problems: Definition, formulas, solved examples and a quiz with practice questions.
Definition
Permutations are the different ways in which a collection of items can be arranged.
For example:
The different ways in which the alphabets A, B and C can be grouped together, taken all at a time, are ABC, ACB, BCA, CBA, CAB, BAC.
Note that ABC and CBA are not same as the order of arrangement is different. The same rule applies while solving any problem in Permutations.
The number of ways in which n things can be arranged, taken all at a time, nPn = n!, called ‘n factorial.’
Factorial Formula
Factorial of a number n is defined as the product of all the numbers from n to 1.
For example, the factorial of 5, 5! = 5*4*3*2*1 = 120.
Therefore, the number of ways in which the 3 letters can be arranged, taken all a time, is 3! = 3*2*1 = 6 ways.
Number of permutations of n things, taken r at a time, denoted by:
nPr = n! / (n-r)!
For example:
The different ways in which the 3 letters, taken 2 at a time, can be arranged is 3!/(3-2)! = 3!/1! = 6 ways.
Important Permutation Formulas
1! = 1
0! = 1
Let us take a look at some examples:
Problem 1: Find the number of words, with or without meaning, that can be formed with the letters of the word ‘CHAIR’.
Solution:
‘CHAIR’ contains 5 letters.
Therefore, the number of words that can be formed with these 5 letters = 5! = 5*4*3*2*1 = 120.
Problem 2: Find the number of words, with or without meaning, that can be formed with the letters of the word ‘INDIA’.
Solution:
The word ‘INDIA’ contains 5 letters and ‘I’ comes twice.
When a letter occurs more than once in a word, we divide the factorial of the number of all letters in the word by the number of occurrences of each letter.
Therefore, the number of words formed by ‘INDIA’ = 5!/2! = 60.
Problem 3: Find the number of words, with or without meaning, that can be formed with the letters of the word ‘SWIMMING?
Solution:
The word ‘SWIMMING contains 8 letters. Of which, I occurs twice and M occurs twice.
Therefore, the number of words formed by this word = 8! / (2!*2!) = 10080.
Problem 4: How many different words can be formed with the letters of the word ‘SUPER’ such that the vowels always come together?
Solution:
The word ‘SUPER’ contains 5 letters.
In order to find the number of permutations that can be formed where the two vowels U and E come together.
In these cases, we group the letters that should come together and consider that group as one letter.
So, the letters are S,P,R, (UE). Now the number of words are 4.
Therefore, the number of ways in which 4 letters can be arranged is 4!
In U and E, the number of ways in which U and E can be arranged is 2!
Hence, the total number of ways in which the letters of the ‘SUPER’ 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
‘BUTTER’ so that the vowels are always together.
Solution:
The word ‘BUTTER’ contains 6 letters.
The letters U and E should always come together. So the letters are B, T, T, R, (UE).
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 ‘REMAINS’ such that the vowels always occur in odd places.
Solution:
The word ‘REMAINS’ has 7 letters.
There are 4 consonants and 3 vowels in it.
Writing in the following way makes it easier to solve these type of questions.
(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.
Combinations
Definition
The different selections possible from a collection of items are called combinations.
For example:
The different selections possible from the alphabets A, B, C, taken 2 at a time, are AB, BC and CA.
It does not matter whether we select A after B or B after A. 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)!]
For example, verifying the above example, the different selections possible from the alphabets A, B, C, taken two at a time are
3C2 = 3! / (2! * (3-2)!) = 3 possible selections (i.e., AB, BC, CA)
Important Combination formulas
nCn = 1
nC0 = 1
nC1 = n
nCr = nC(n-r)
The number of selections possible with A, B, C, taken all at a time is 3C3 = 1 (i.e. ABC)
Solved examples of Combination
Let us take a look at some examples to understand how Combinations work:
Problem 1: In
how many ways can a committee of 1 man and 3 women can be formed from a group of 3 men and 4 women?
Solution:
No. of ways 1 man can be selected from a group of 3 men = 3C1 = 3! / 1!*(3-1)! = 3 ways.
No. of ways 3 women can be selected from a group of 4 women = 4C3 = 4! / (3!*1!) = 4 ways.
Problem 2: Among a set of 5 black balls and 3 red balls, how many selections of 5 balls can be made such that at least 3 of them are black balls.
Solution:
Selecting at least 3 black balls from a set of 5 black balls in a total selection of 5 balls can be
3 B and 2 R
4 B and 1 R and
5 B and 0 R balls.
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 3, 5, 7, 8, 9, 0 such that no number repeats?
Solution:
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.
Permutations and Combinations Quiz
Try these practice problems.
Problem 1: Click here
Answer 1: Click here
Problem 2: Click here
Answer 2: Click here
Problem 3: Click here
Answer 3: Click here