Question

The number of times the digit 5 will be written when listing the integers from 1 to 1000 is: -
(a) 250
(b) 300
(c) 350
(d) 400

Answer 1

Hint:
Consider three boxes in which the digit 5 is to be filled. Now, consider three cases, in the first case select one place by using the relation \[{}^{3}{{C}_{1}}\] and find the possible ways to fill the digit 5 in this selected box and digits other than 5 in the remaining two boxes. In the second case select two boxes using the relation \[{}^{3}{{C}_{2}}\] and fill 5 in them then find the possible ways to fill the remaining box with the digits other than 5. In case three find the number of ways to fill three 5’s in all the three boxes. Find the number of 5’s appearing in each case and add them to get the answer.

Complete step by step answer:
Here, we have been provided with the numbers between 1 and 1000 including them also and we have been asked to find the number of times the digit 5 will be written.
Now, let us assume three boxes in which the digit 5 will be filled. So, we will have the following three cases: -
1. Case (1): - When the digit 5 is present only once in a number. So, we have,

Here, we have three places and only one needs to be selected to get it filled with 5. So, we have,
\[\Rightarrow \] Number of ways to select one place from the given three places will be = \[{}^{3}{{C}_{1}}\] = 3
Now, the other two boxes must be filled with digits other than 5, so we have 9 digits left. Therefore, we get,
\[\Rightarrow \] Number of ways to fill the other two places = 9 \[\times \] 9 = 81
So, number of different numbers in which the digit 5 appear once = 3 \[\times \] 81 = 243
Now, we can see that in each number the digit 5 will appear one time, therefore,
\[\Rightarrow \] Number of times the digit 5 will be written in this case = 1 \[\times \] 243 = 243
2. Case (2): - When the digit 5 is present at two places in a number. So, we have,

Here, we have three places and two of them need to be filled with 5. So, we have,
\[\Rightarrow \] Number of ways to select two places from the given three places will be = \[{}^{3}{{C}_{2}}\] = 3
Now, the remaining one box must be filled with digits other than 5, so we have 9 digits left. Therefore, we get,
\[\Rightarrow \] Number of ways to fill the remaining one box = 9
So, number of different numbers in which the digit appear twice = 3 \[\times \] 9 = 27
Now, we can see that in each number the digit 5 will appear two times, therefore,
\[\Rightarrow \] Number of times the digit 5 will be written in this case = 2 \[\times \] 27 = 54
3. Case (3): - When the digit 5 is present at all the three places in a number. So, we have,

5

5

5

Clearly, we can see that there is only one such number, i.e. 555, in which all the three boxes are filled with the digit 5.
Now, we can see that in 555 the digit 5 will appear three times, therefore,
\[\Rightarrow \] Number of times the digit 5 will be written in this case = 1 \[\times \] 3 = 3
So, considering all the three cases, we get,
\[\Rightarrow \] The number of times the digit 5 will be written when listing the integers from 1 to 1000 = 243 + 54 + 3 = 300.
Hence, option (b) is the correct answer.
Note:
 One may note that we have considered only 3 boxes because the numbers are from 1 to 1000 and if we will consider a \[{{4}^{th}}\] box and fill it with 5 then the number will become greater than 1000 and our condition will become false. You must remember the formula of combination which is used for selecting objects. Note that here you do not have to apply the formula for permutation of objects because it is used for the arrangement of objects and not selection.