# 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 â€˜LUCKNOWâ€™ which have

- no restriction
- L as the 1st letter of the word
- L and W as the terminal letters
- all the vowels together
- L always occuring before U
- L always occuring before U and U always occuring before W

Solution

- The total number of distinct letters = 7 (L, U, C, K, N, O, W)
- Now we can arrange only 6 letters (as place of L is restricted),
- Now we can arrange only 5 letters (as place of L and W are restricted),
- U and O should be together, so we will assume these two letters to be tied up with each other.
- 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.
- 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.

Example-2

How many new words can be formed with the word â€˜PATNAâ€™?

Solution

Total number of lettersâ€” 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 â€“1 = 59

Example-3

How many words can be formed with the word â€˜ALLAHABADâ€™?

Solution

Letters areâ€” A â€“ Four times

L â€“ Twice

H, B and D occur once.

So, the total number of words =

Example-4

How many 4-lettered distinct words can be formed from the letters of the word â€˜EXAMINATIONâ€™?

Solution

Letters are:A â€“ Twice

I â€“ Twice

N â€“ Twice

E, X, M, T, O â€“ Once

I â€“ Twice

N â€“ Twice

E, X, M, T, O â€“ 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.**- All distinct =
^{8}P_{4}(Distinct letters are â€“ A, I, N, E, X, M, T and O) - Two same, two distinct
^{3}C_{1}ways. Now rest of the two distinct letters can be chosen in^{7}C_{2}^{ }ways.^{3}C_{1}Ã—^{7}C_{2}Ã— - 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
^{3}C_{2}^{ }ways.^{3}C_{2}Ã—

# 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

If a three-digit number is to be constructed, then we will use

Hundredâ€™s place | Tenâ€™s place | Unitâ€™s place |

If a four-digit number is to be constructed, then we will use and so on.

Thousandâ€™s place | Hundredâ€™s place | Tenâ€™s place | Unitâ€™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?

- When repetition is not allowed
- When repetition is allowed

Solution

The box given below represents the respective positioning of digits in a three-digit number.

Hundredâ€™s place | Tenâ€™s place | Unitâ€™s place |

- Since repetition of the digits is not allowed, we can fill the unitâ€™s place in 5 ways, tenâ€™s place in 4 ways and hundredâ€™s place in 3 ways.

Using Multiplication Theorem, the total number of numbers which can be formed = 5 Ã— 4 Ã— 3 = 60
Alternatively, 3 digits can be selected out of 5 digits in

^{5}P_{3}= 60- Since repetition of the digits is allowed here, we can fill each of the hundredâ€™s, tenâ€™s and unitâ€™s place in 5 ways.

Using Multiplication Theorem, the total number of numbers which can be formed = 5 Ã— 5 Ã— 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 hundredâ€™s and thousandâ€™s place:
Thousandâ€™s place cannot be filled by 0, so it can be filled in 5 ways.
Hundredâ€™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 Ã— 6 = 30
So, the total number of numbers which can be formed = 30 Ã— 9 = 270

In this case, last two digits of the number can be 00, 04, 12, 20, 24, 32, 40, 44, 52.

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 beâ€”04, 12, 20, 24, 32, 40, 52.

At this point, we will have to divide the process of solving this questionâ€”one part will have those numbers which containâ€˜0â€™ 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 â€“ Last two digits are 04, 20, 40.

= 4 Ã— 3 = 12 ways

Hence, the total number of numbers which can be formed = 12 Ã— 3 = 36
2nd part â€” Last two digits are 12, 24, 32, 52.

â€˜0â€™ cannot occur at thousandâ€™s place.

= 3 Ã— 3 = 9 ways

Hence, the total number of numbers which can be formed = 9 Ã— 4 = 36

Total numbers = 36 + 36 = 72

At this point, we will have to divide the process of solving this questionâ€”one part will have those numbers which containâ€˜0â€™ as one of its last two digits viz., 04, 20, 40 and other part will have the remaining numbers viz., 12, 24, 32, 52.

= 4 Ã— 3 = 12 ways

Hence, the total number of numbers which can be formed = 12 Ã— 3 = 36

â€˜0â€™ cannot occur at thousandâ€™s place.

= 3 Ã— 3 = 9 ways

Hence, the total number of numbers which can be formed = 9 Ã— 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:

questions:

- For a number to be odd, unit digit should be either 1 or 3 or 5 or 7 or 9.
- Thousandâ€™s place cannot be filled with 8 or 9.

For unitâ€™s digit â€” When it is filled with 9, thousandâ€™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 Ã— 7 = 56 ways.
So, the total number of numbers formed in this way = 56 Ã— 7 = 392
Now, if the unitâ€™s place is filled with any of the four digits 1, 3, 5 or 7, the thousandâ€™s place can be filled in 6 ways (0 will be excluded), and the remaining two places can be filled in 8 Ã— 7 = 56 ways.
So, the total number of numbers formed in this way = 56 Ã— 6 Ã— 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 Ã— 9 Ã— 8 Ã— 7 = 504
So, the numbers having atleast one of its digits repeated = 1000 â€“ 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,
= 2 Ã— 3 = 6 numbers
For three-digit numbers,
= 2 Ã— 3 Ã— 3 = 18 numbers
For four-digit numbers,an
= 2 Ã— 3 Ã— 3 Ã— 3 = 54 numbers
For five-digit numbers,
= 2 Ã— 3 Ã— 3 Ã— 3 Ã— 3 = 162 numbers
For six-digit numbers,
= 2 Ã— 3 Ã— 3 Ã— 3 Ã— 3 Ã— 3 = 486 numbers
So, the total number of numbers = 728