Question

Suppose the six digits 1, 2, 4, 5, 6, 7 are given to us and we have to find the total number of numbers with no repetition of digits which can be formed under different conditions.

Answer 1

1. There is no restriction. The number of 6-digit numbers. 

= ⁶P₆ = 6 ! = 6 × 5 × 4 × 3 × 2 × 1= 720. 

2. Numbers in which a particular digit occupies a particular place. Suppose we have to form numbers in which 5 always occurs in the ten’s place. In this case, the ten’s place is fixed and the remaining five places can be filled in by the remaining 5 digits in ⁵P₅, i.e., 5 ! = 120 ways. 

The number of numbers in which 5 occurs in the ten’s place = 120. 

3. Numbers divisible by a particular number. Suppose we have to form numbers which may be divisible by 2. These numbers will have 2 or 4 or 6 in the unit’s place. Thus the unit’s place can be filled in 3 ways. After having filled up the unit’s place in any one of the above ways, the remaining five places can be filled in ⁵P₅ = 5 ! = 120 ways. 

∴ The total number of numbers divisible by 2 = 120 × 3 = 360. 

4. Numbers having particular digits in the beginning and the end. Suppose we have to form numbers which begin with 1 and end with 5. Here, the first and the last places are fixed and the remaining four places can be filled in 4 !, i.e., 24 ways by the remaining four digits. Therefore, the total number of numbers beginning with 1 and ending with 5 = 24. 

Note. If the numbers could have 1 or 5 in the beginning or the end, the number would have been 2 ! . 4 !, i.e., 48. 

5. Numbers which are smaller than or greater than a particular number. Suppose we have to form numbers which are greater than 4,00,000. In these numbers, there will be 4 or a digit greater than 4, i.e. 5, 6 or 7 in the lac’s place. Thus this place can be filled in 4 ways. The remaining 5 places can then be filled in 5 ! = 120 ways. 

∴ The total number of numbers = 4 × 120 = 480.