Question

The number of positive integers ≤100000 which contain exactly one 2, one 5 and 7 in its decimal representation is

Answer 1

Hint: Using proper logic we can solve a given problem. The decimal representation of a rational number cannot be non-terminating non-repeating, as the decimal expansion of rational number is either terminating or non-terminating repeating .

Complete step-by-step answer:
The number of 5-digit numbers (where the leading digits can be 0) which contain exactly one 2, one 5, and one 7.
Start with the number of 5-digit numbers 257XY, where X and Y are not 2, 5, or 7.
There are 7 choices for X (0,1,3,4,6,8, and 9) and the same 7 choices for Y.
 $$7 \times 7{\text{ }} = {\text{ }}49$$.
How many ways can each 257XY be rearranged?
There are 5 choices for the first digit, and then 4 choices remain for the second digit, etc.
$$5!{\text{ }} = {\text{ }}120$$
How many total arrangements is that?
$$49{\text{ }} \times 120{\text{ }} = {\text{ }}5880$$
Have we counted every number that fits the requirement?
Yes. For example, 752 is an arrangement of 25700.
Have we counted any number twice?
Yes. In fact we've counted every number twice. Arranging 25701 gives all the same numbers as arranging 25710. Arranging 25733 produces 23537 and 23537.
$$\dfrac{{5880}}{2} = 2940$$

Note: The rational numbers can be represented in the form of decimals rather than in the form of fractions. A rational number can be expressed as a terminating or non terminating, recurring decimal.