Question

The total number of times, the digit 3 will be written when the integers having less than 4 digits are listed is equal to

& A.300 \\\ & B.310 \\\ & C.302 \\\ & D.306 \\\ \end{aligned}$$

Answer 1

In this question, we have to find the number of times digit 3 will be written when integers less than 4 digits are listed. For this, we will find the number of times 3 can be repeated giving us three ways numbers can be written. After this we will find the count of numbers of each type and hence add them to get the required answer. We will use combination formula given by ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ where r things are chosen from n items.

Complete step by step answer:
Let us solve this question, considering every possibility. First, let’s see what type of numbers can be listed. As we have to list numbers less than 4 digit, so numbers will be selected from 0 to 999 and since 3 digits is counted, so numbers can have three forms: (1) having one three (2) having two three's and (3) having three threes.
Now, let’s count possibilities for all three cases one by one. We will consider three places for every case because numbers should be taken less than four digits and zeros can make one digit and two digit numbers:
(1) When only one 3 appears.
In this case, there are three places and only one 3 is to be placed ${}^{3}{{C}_{1}}$. Now for the other two places, there are remaining 9 digits (0, 1, 2, 4, 5, 6, 7, 8, 9) from which two of the places will be covered. So ways of doing that becomes $9\times 9$. Also 3 is coming 1 time, so it will be counted one time only.
Hence, total ways for one 3 appears is ${}^{3}{{C}_{1}}\times 9\times 9\times 1=3\times 9\times 9=243$.
(2) When only two 3 appear.
In this two 3's will take place in three places, so the number of ways for selecting a place becomes ${}^{3}{{C}_{2}}$. For the remaining position, there are 9 possible ways left. And at last 3 are coming twice, so it will be 2 times. Hence, total ways for two 3 appears is ${}^{3}{{C}_{2}}\times 9\times 2=3\times 9\times 2=54$.
(3) When all places are covered by three, there is only one number but 3 will be counted 3 times. So the number of ways for three 3's to be written is 3.
Using all three parts, the total number of times 3 will be counted becomes equal to $243+54+3=300$.
Hence, 300 is our required answer.

So, the correct answer is “Option A”.

Note: In this question, students can make the mistake of counting only the numbers which contain 3 and forget about counting the number of 3’s. for numbers having two threes, we will count that number twice and for numbers having three three’s we will count that number thrice. Students should take care while using the formula ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. Here, r can never be greater than n.