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Word Formation

As we know that order of occurrence of letters decide the formation of words, so word formation is one standard example of permutation.
 
Let us understand with the help of some examples:
 
Example-1
How many words can be formed with the word ‘LUCKNOW’ which have
  1. no restriction
  2. L as the 1st letter of the word
  3. L and W as the terminal letters
  4. all the vowels together
  5. L always occuring before U
  6. L always occuring before U and U always occuring before W
Solution
  1. The total number of distinct letters = 7 (L, U, C, K, N, O, W)
     
    So, the total number of words that can be formed = 7!
  2. Now we can arrange only 6 letters (as place of L is restricted),
     
    So, the total number of words that can be formed = 6!
  3. Now we can arrange only 5 letters (as place of L and W are restricted),
     
    So, the number of arrangements = 5!
     
    But the place of L and W can be interchanged between themselves.
     
    So, the total number of words that can be formed = 5! × 2!
  4. U and O should be together, so we will assume these two letters to be tied up with each other.
     
    Now we have 6 distinct things to be arranged– (L, U, O, C, K, N, W)
     
    So, the number of arrangements = 6!
     
    But place of U and O can be interchanged between themselves.
     
    So, the total number of words that can be formed = 6! × 2!
  5. There is an equal likelihood occurrence of all the letters in the word, so in half of the cases L will occur before U and in the remaining half, U will occur before O.
     
    So, the total number of words that can be formed = 7!/2
  6. There are six possible arrangements (3!) corresponding to L, U and W. However, only one out of these six will be in the prescribed order: L always occurs before U and U always occurs before W.
     
    So, corresponding to 7! arrangements, the number of ways in which the condition will be satisfied = 7!/3! ways.
 
Example-2
How many new words can be formed with the word ‘PATNA’?
Solution
Total number of letters— P, T, N occur once while A occurs twice.
So, the total number of words that can be formed
= 5!/2! = 60
Total number of new words = 60 –1 = 59
 
Example-3
How many words can be formed with the word ‘ALLAHABAD’?
Solution
Letters are— A – Four times
L – Twice
H, B and D occur once.
So, the total number of words = Description: 2205.png
 
 
Example-4
How many 4-lettered distinct words can be formed from the letters of the word ‘EXAMINATION’?
Solution
Letters are:A – Twice
I – Twice
N – Twice
E, X, M, T, O – Once
 
 
Words will be of three types:
 
i. All distinct, ii. Two same, two distinct, iii. Two same and of one kind: two same and of other kind.
  1. All distinct = 8P4 (Distinct letters are – A, I, N, E, X, M, T and O)
  2. Two same, two distinct
     
    Selection of one pair out of the three pairs (A, I, N) can be chosen in 3C1 ways. Now rest of the two distinct letters can be chosen in 7C2 ways.
     
    Total number of words = 3C1 × 7C2 × Description: 2214.png
  3. Two same and are of one kind, two same and are of other kind = Out of the three pairs of letters (A, I, N), we can select two pairs in 3C2 ways.
     
    Total number of words = 3C2 × Description: 2223.png

Number Formation

Number formation is another standard example of permutation. Here we will discuss the box diagram method of solving the questions.

If a three-digit number is to be constructed, then we will use
 
Hundred’s place Ten’s place Unit’s place
 
If a four-digit number is to be constructed, then we will use and so on.
 
Thousand’s place Hundred’s place Ten’s place Unit’s place

While solving the questions related to number formation,
 we should know two things very clearly:
  • While using the box diagram, we should start with the digit which has restriction, i.e., some condition is imposed on that digit.
  • When nothing about the repetition of digits is mentioned in the question, we will assume that the repetition is allowed.
Example-1
How many different 3-digit numbers can be formed using the digits 1, 2, 3, 4 and 5?
  1. When repetition is not allowed
  2. When repetition is allowed
Solution
The box given below represents the respective positioning of digits in a three-digit number.
Hundred’s place Ten’s place Unit’s place
  1. Since repetition of the digits is not allowed, we can fill the unit’s place in 5 ways, ten’s place in 4 ways and hundred’s place in 3 ways.
Description: 2233.png 
Using Multiplication Theorem, the total number of numbers which can be formed = 5 × 4 × 3 = 60
 
Alternatively, 3 digits can be selected out of 5 digits in 5P3 = 60
  1. Since repetition of the digits is allowed here, we can fill each of the hundred’s, ten’s and unit’s place in 5 ways.
Description: 2242.png
 
Using Multiplication Theorem, the total number of numbers which can be formed = 5 × 5 × 5 = 125
 

Example-2
How many 4-lettered numbers divisible by 4 can be formed from the digits 0, 1, 2, 3, 4, 5?
Solution
Any number divisible by 4 will have the number formed by its last two digits divisible by 4.
In this case, last two digits of the number can be 00, 04, 12, 20, 24, 32, 40, 44, 52.
 
Corresponding to any one of 00, 04, 12, 20, 24, 32, 40, 44, 52, we can have the following digits at its hundred’s and thousand’s place:
Description: 2251.png 
 
Thousand’s place cannot be filled by 0, so it can be filled in 5 ways.
 
Hundred’s place can be filled by any of the 0, 1, 2, 3, 4, 5; hence 6 ways.
 
So, corresponding to any one of 00, 04, 12, 20, 24, 32, 40, 44, 52, the total number of ways = 5 × 6 = 30
 
So, the total number of numbers which can be formed = 30 × 9 = 270
 

Example-3
In the above question, how many numbers can be formed if repetition of the digits is not allowed?
Solution
Last two digits of this number can be—04, 12, 20, 24, 32, 40, 52.
At this point, we will have to divide the process of solving this question—one part will have those numbers which contain‘0’ as one of its last two digits viz., 04, 20, 40 and other part will have the remaining numbers viz., 12, 24, 32, 52.
 
1st part – Last two digits are 04, 20, 40.
Description: 2260.png 
= 4 × 3 = 12 ways
Hence, the total number of numbers which can be formed = 12 × 3 = 36
 
2nd part — Last two digits are 12, 24, 32, 52.
‘0’ cannot occur at thousand’s place.
Description: 2269.png 
= 3 × 3 = 9 ways
Hence, the total number of numbers which can be formed = 9 × 4 = 36
Total numbers = 36 + 36 = 72
 

Example-4
How many odd integers from 1000 to 8000 have none of its digits repeated?
Solution
There are two restrictions operating in this
questions:
  1. For a number to be odd, unit digit should be either 1 or 3 or 5 or 7 or 9.
  2. Thousand’s place cannot be filled with 8 or 9.
For unit’s digit — When it is filled with 9, thousand’s place can be filled in 7 ways namely any digit from 1 to 7, and the remaining two places can be filled in 8 × 7 = 56 ways.
 
So, the total number of numbers formed in this way = 56 × 7 = 392
 
Now, if the unit’s place is filled with any of the four digits 1, 3, 5 or 7, the thousand’s place can be filled in 6 ways (0 will be excluded), and the remaining two places can be filled in 8 × 7 = 56 ways.
 
So, the total number of numbers formed in this way = 56 × 6 × 4 = 1344
 
So, the total number of numbers = 392 + 1344 = 1736
 

Example-5
How many integers from 6000 to 6999 have atleast one of its digits repeated?
Solution
Total number of numbers = None of its digits repeated numbers + atleast one of its digits repeated number (i.e., either the digits will be repeated or not repeated).
 
Total numbers with none of its digits repeated = 1 × 9 × 8 × 7 = 504
 
So, the numbers having atleast one of its digits repeated = 1000 – 504 = 496
 
 
Example-6
How many natural numbers less than a million can be formed using the digits 0, 7 and 8?
Solution
The numbers formed would be of a single digit, two digits, three digits, four digits, five digits and six digits.
 
Single-digit numbers = 7 and 8
 
For two-digit numbers,
Description: 2278.png
= 2 × 3 = 6 numbers
 
For three-digit numbers,
Description: 2287.png
= 2 × 3 × 3 = 18 numbers
 
For four-digit numbers,an
Description: 2296.png
= 2 × 3 × 3 × 3 = 54 numbers
 
For five-digit numbers,
Description: 2305.png
= 2 × 3 × 3 × 3 × 3 = 162 numbers
 
For six-digit numbers,
Description: 2314.png
= 2 × 3 × 3 × 3 × 3 × 3 = 486 numbers
 
So, the total number of numbers = 728
 





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