Find a smallest four digit number using the digits 7, 8, 9, 5 such that the number thus formed has 9 at its hundreds place and 8 at its one’s place.
A. 5978
B. 7598
C. 7958
D. 7985

Hint: Form a number using these four digits such that ones and hundreds place of the number contain the same as given in question and arrange the remaining digits in the remaining two places in such a way that the formed number is as small as possible.

Complete step-by-step answer:
Given digits are 7, 8, 9, 5. We have two form a four - digits number using digits 7, 8, 9 and 5.
In a four number, there are four places;
1. Ones
2. Tens
3. Hundreds
4. Thousands
First place from the right is called one.
Second place from the right is called tens.
Third place from right is called hundreds
Fourth place from right is called thousands.
For example, let’s consider a four digit number “9856”.
‘6’ is at one’s place, ‘5’ is at tens place, ‘8’ is at hundred’s place and ‘9’ is at thousands place.
In our question, it is mentioned that ‘9’ is at hundreds place and ‘8’ is at ones place i.e. $\_9\_8$
Now, two places are remaining and two digits are also remaining i.e. 5 and 7.
We have to arrange these two digits in the remaining two places.
There are two possible numbers which can be formed;
1. 5978
2. 7958
But in question it is mentioned that we want the smallest number. So, the smallest number out of these two numbers will be our answer.
So, option a) 5978 is the correct answer.

Note: The possibility of making a mistake is by not considering the possible two cases of 5978 and 7958. We have to consider both the cases and take the smallest one as an answer. We consider the digit’s places from the right side to form a number.