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## Comment on

2 Repeated Digits## How to solve it using ignore

## (# with exactly 2 repeats) =

(# with exactly 2 repeats) = (number of integers from 500 to 999 inclusive) - (# with zero repeats) - (# with 3 repeats)

- Number of integers from 500 to 999 inclusive = 999-500+1 = 500

- # with zero repeats = (5)(9)(8) = 360 because first digit can be 5,6,7,8, or 9. 2nd digit can any digit other than first digit. 3rd digit can any digit other than first two digits.

- # with 3 repeats = 5 (555, 666, 777, 888 and 999)

So, # with exactly 2 repeats = 500 - 360 - 5 = 135

## Can I solve this using the

I got the following logic

# ways to get first digit = 5

# ways to get second digit = 9

# ways to get third digit = 2 (it can either be equal to 1 or 2)

By this logic, I get answer as 5 x 9 x 2 = 90.

Having difficulty getting 135 by a combined logic

## Hi Mohit,

Hi Mohit,

Your solution covers two possible cases:

Case 1) The 1st and last digits are the same (e.g., 626 and 747)

Case 2) The 2nd and last digits are the same (e.g., 677 and 800)

However, you haven't considered....

Case 3) The 1st and 2nd digits are the same (e.g., 557 and 992)

In how many ways can we satisfy case 3?

# ways to get first digit = 5

# ways to get second digit = 1 (must match 1st digit)

# ways to get third digit = 9 (can be ANY digit other than the 1st and 2nd digit)

We get: 5 x 1 x 9 = 45

You already got a total of 90 outcomes for cases 1 and 2.

So, when we add that to the 45 outcomes for case 3, we get 135

Cheers,

Brent

## I approached this by saying

I then went through the process of finding the number of numbers from 500-999 that have only unique digits

5*9*8 = 360

Numbers from 500-999 that contain 3 equal digits

5*1*1 (originally forgot this step and get 140 instead of 135)

500 -360 - 5 = 135

Is this a valid approach?

## That's a perfectly valid

That's a perfectly valid approach. Nice work!