Hi,

Having this exercise:

How many integers between 100 and 999 inclusive have repeated digits?

I have resolved using this reasoning, I think correct, distinguishing four different cases:

case I)

three repeated digits:

the first digit can be one of the possible 9 numbers in the set {1,2,...,9},

the second and the third digits must be the same number of the first digit:

9·1·1 = 9

case II)

the FIRST two digits are repeated,

in the third digit also the zero is included, but, since we have used one number for the previous digits, we have 10-1=9 possible numbers

9·1·9 = 81

case III)

the LAST two digits are repeated,

in the second digit also the zero is included, but, since we have used one number for the previous digit, we have 10-1=9 possible numbers

9·9·1 = 81

case IV)

the first and the third digits are repeated:

same of case iii)

9·9·1 = 81

total:

9 + 3(81) = 9 + 243 = 252

Please, I would to know, is there any alternative method where it is not necessary to consider 4 different cases?

(In fact if you consider a larger integer, the distinction of different cases seems to be a lot of work. )

many thanks!

## How many integers between 100 and 999 with repeated digits

**Moderators:** gmalivuk, Moderators General, Prelates

### Re: How many integers between 100 and 999 with repeated digi

How many integers do not have repeating digits?

9 choices for the first digit (no zero), 9 for the second (no repetition), 8 for the third => 9*9*8=648

648+252=900 as expected.

9 choices for the first digit (no zero), 9 for the second (no repetition), 8 for the third => 9*9*8=648

648+252=900 as expected.

- mathmannix
**Posts:**1446**Joined:**Fri Jul 06, 2012 2:12 pm UTC**Location:**Washington, DC

### Re: How many integers between 100 and 999 with repeated digi

mfb has the best solution.

So, while not really another way, you could also just kind of collapse your cases II-IV into one case...

New Case 2:

How many integers have two repeated digits?

9 · 9 · 1 ·

So, while not really another way, you could also just kind of collapse your cases II-IV into one case...

New Case 2:

How many integers have two repeated digits?

9 · 9 · 1 ·

_{3}C_{2}= 243I hear velociraptor tastes like chicken.

### Re: How many integers between 100 and 999 with repeated digi

Or with two cases instead of 4: The two cases being the second digit is identical to first: 9 (no zero)*1 (identical to the first)*10 (can be anything). And the second case that it isn't the same as the first: 9*9(anything beside the first)*2(one of the first two.)

9*1*10+9*9*2=252

Though yes going the other way seem to be the elegant option.

9*1*10+9*9*2=252

Though yes going the other way seem to be the elegant option.

### Re: How many integers between 100 and 999 with repeated digi

many thanks!

Even if it's trivial, can you explain me better this passage?

mathmannix wrote:So, while not really another way, you could also just kind of collapse your cases II-IV into one case...

New Case 2:

How many integers have two repeated digits?

9 · 9 · 1 ·_{3}C_{2}= 243

Even if it's trivial, can you explain me better this passage?

### Re: How many integers between 100 and 999 with repeated digi

effbert wrote:many thanks!mathmannix wrote:So, while not really another way, you could also just kind of collapse your cases II-IV into one case...

New Case 2:

How many integers have two repeated digits?

9 · 9 · 1 ·_{3}C_{2}= 243

Even if it's trivial, can you explain me better this passage?

There are 9 ways to choose the first digit (not zero). There are 9 ways to choose a different digit for one of the latter digits. There are three ways to choose which of the three digits is unique (aka

_{3}C

_{2}). This covers all 3 cases in a single expression.

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