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Case1: _ _ 7 Digits 0 - 9 can take first place in 10 ways. Similarly 0 - 9 digits can fill up the second place in 10 ways. Total ways = m * n = 10 * 10 = 100

since there are 10 digits and repetitions can be allowed . keeping 7 in hundreds place >> 7 __ __ { 1*10*10} keeping 7 in tens place >> __ 7[u]__ { 10*1*10} keeping 7 in units place >> __ __ [u]7 {10*10*1} adding above 3 we get 300

is this approach correct there are 3 places x,x,x consider 7 can be placed in each of the position case 1 7 at hundredths place then rest 2 position can be filled by 10 numbers (0-9) there fore total in 100 ways (1X10X10) (possibilities are in and clause) case2 7 at tenths place then rest 2 position can be filled by 10 numbers (0-9) there fore total in 100 ways (10X1X10) (possibilities are in and clause) case 3 7 at ones place then rest 2 position can be filled by 10 numbers (0-9) there fore total in 100 ways (10X10X1) (possibilities are in and clause) as these possibilities are in or clause so adding them up gives 300 ways

Haihai, I believe that's because when you have four digits the numbers could exceed 1000 such as 7,674 and so on. I believe what you did would be the approach to finding out how many times 7 is used between 1 and 10,000.

I think this is a high-quality question and the explanation isn't clear enough, please elaborate. Consider numbers from 1 to 1000 written as follows:

1. 0001

2. 0002

3. 0003

...

1000. 1000

We still have 1000 numbers. However, we used 4 digits per number, hence used total of 4∗1000=4000 digits. 4000/10=400 times.

What's wrong with the above inference? Thanks.

It doesn't work because you have 4 digits (and aren't accounting for 1001 to 9999). The concept applies as follows:

1 digit (0 to 9): 10 numbers, 1 digit per number = 10 digits --> 10/(10 digits) = each digit used 1 time 2 digits (00 to 99): 10*10=100 numbers, 2 digits per number = 200 digits --> 200/(10 digits) = each digit used 20 times 3 digits (000 to 999): 10*10*10=1000 numbers, 3 digits per number = 3000 digits --> 3000/(10 digits) = each digit used 300 times 4 digits (0000 to 9999): 10*10*10*10=10000 numbers, 4 digits per number = 40000 digits --> 40000/(10 digits) = each digit used 4000 times .. n digits: 10^n numbers, n digits per number = n*10^n digits --> (n*10^n)/10 = each digit used n*10^(n-1) times

POSSIBILE COMBO NUMBER OF 7's use X as every single digit number except 7 -->X=0,1,2,3,4,5,6,8,9 --> 9 POSSIBLE NUMBERS 7XX 81 XX7 81 X7X 81 X77 9x2 77X 9x2 7X7 9x2 777 3 _____ 300

For a 3 digit number, 7 can be at the unit place, tens place or hundred place. 7 at hundred place, that means total combination will be:- 1*10*10 (first digit can only be seven but second and third can be any from 0-9) 7 at tens place, that means total combination will be:- 9*1*10 (first digit can be any number from 1-9, second will only be 7 and third will be any number from 0-9) 7 at unit place, that means total combination will be:- 9*10*1 (first digit can be any number from 1-9, second will any number from 0-9 and third will be 7) So total 3 digit number with 7 at some place = 100+90+90=280 (You can calculate this for any digit)

For a 2 digit number, 7 can be at unit place, or tens place Going with the same logic as described above:- 7 at tens place total combination will be:- 1*10 (first digit can only be seven but second can be any from 0-9) 7 at unit place total combination will be:- 9*1 (first digit can be any number from 1-9 and second digit is 7) So for 2 digit number total combination will be 10+9=19 (You can calculate this for any digit)

For single digit number, there is only one possibility So total is 280+19+1=300. D is the answer.

X X 7 = 9*9*1 X 7 X = 9*1*9 7 X X = 1*9*9 7 7 X = 1*1*9 7 X 7 = 1*9*1 X 7 7 = 9*1*1 777 = 1 if we add all of them = 81+81+81+9+9+9+1=271 where did i go wrong??

in cases 77X, 7X7, X77 and 777 the digit appears 1*1*9 * 2, 1*9*1*2, 9*1*1*2, 1*3. You do this the answer is exactly 300.

Note- Question is about number of 7, not the numbers which have seven. eg. 777 has 3 sevens.

I think this is a high-quality question and I don't agree with the explanation.

Hi amar.igtr,

Thank you for posting. One request though: please be specific, why don't you agree with the solution? So, far you've posted two similar replies and none of them were correct. Thank you.
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