How many times will the digit 7 be written when listing the integers from 1 to 2000?

269308300None of these

Answer : C

Solution : Before 1000 there are one digit, two digits and three digits numbers.
Number of times 3 appear in one digit number `= 20 xx 9`
Number of times 3 appear in two digit number `= 11 xx 9`
Number of times 3 appear in three digit number = 21
Hence total number of times the digit 3 appear while writing the integers from 1 to 1000
`= 180 + 99 + 21 = 300`

We are asked to find the number of times 5 appears from 1 to 1000.

Now, in 1000 there are no 5 digits so 5 appears in one digit, two digits, and three digits numbers.

We are going to find the number of times 5 appears from 1 to 1000 by taking three cases. In the first case, 5 appears once, in the second case 5 appears twice and in the third case, 5 appears thrice and then adds all the three cases.

Case 1: When the digit 5 appears once.

_ _ _

In the three places above, first of all we are going to choose one place where 5 appears. The ways to select one position out of these three places is:

${}^{3}{{C}_{1}}$ 

Now, in the remaining 2 places, any 9 digits from 0 to 9 except 5 will appear so multiplying 9 by 9 and hence multiplying ${{9}^{2}}$ to above expression we get,

${}^{3}{{C}_{1}}{{\left[ 9 \right]}^{2}}$

We know that, ${}^{n}{{C}_{1}}=n$. Using this relation in the above expression we get,

$\begin{align}

  & 3\left[ 81 \right] \\ 

 & =243 \\ 

\end{align}$ 

Case 2: When the digit 5 appears twice.

_ _ _

Now, the digits are appearing twice so we have to choose two places out of three places and the ways to select two places out of 3 are as follows:

${}^{3}{{C}_{2}}$ 

We know that, ${}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}$ using this relation in the above expression we get,

${}^{3}{{C}_{3-2}}={}^{3}{{C}_{1}}$ 

The above expression is reduced to 3.

Now we know that if in a number 5 appears twice then we have to multiply numbers by 2.

After the two places are occupied by 5 we are remaining with just one position so any of the 9 digits can be placed in that blank so multiplying 9 by 3 we get,

$\begin{align}

  & 9\times 3 \times 2\\ 

 & =54 \\ 

\end{align}$ 

Case3: When the digit 5 appears thrice.

Now we know that if in a number 5 appears thrice then we have to multiply numbers by 3.

There is only one possibility when 5 appears thrice is:

$555$ 

So the number will be $ 3 \times 1$.

Now, adding the result of cases 1, 2 and 3 we get,

$\begin{align}

  & 243+54+3 \\ 

 & =300 \\ 

\end{align}$ 

Hence, the digit 5 appears 300 times from 1 to 1000.

Hence, the correct option is [c].

Let us take a systematic approach to answer this [make a list]:

For numbers [3 digit] when there is EXACTLY one 7  there are 9 possibilities [0,1,2,3,4,5,6,8,9] for each of the other 2 digits i.e 81. However the 7 could be in the units, tens or hundreds place so there is a total of 81 x 3 = 243 occurrences.

- - 7;  - 7 -; 7 - -.

For numbers [3 digit] when there are EXACTLY two 7's  there are 9 possibilities for the other digit. However the two 7's could be positioned in 3 ways so there is a total of 9 x 3 = 27 occurrences. 

- 7 7; 7 - 7; 7 7 -

There is exactly 1 number with all three sevens [777] .

So we have a total of 243 + 27 + 1 =271.

Note

ALL the BEST!

Pradip

A. 271

B. 300

C. 252

D. 304