Diversity of Digits

Probability Level 1

How many positive integers less than 101 contain two different digits?

Do not include leading zeros when counting the digits

80 91 82 90

7 solutions

Chung Kevin
Oct 17, 2015

There are 100 numbers in the range 1-100.
The numbers that don't meet the given criteria of using at least two different digits are:
1, 2, 3, 4, 5, 6, 7, 8, 9 and 11, 22, 33, 44, 55, 66, 77, 88, 99

$100-18 = \fbox{82}$

But why are you including 100 into this? It is not a two digit number.

Rajat Rathore - 5 years, 7 months ago

He didn't state that he wanted a "2 digit number" he only needed how many numbers "contain" 2 different digits and 100 contains 1 & 0 which fulfills the requirement

Ahmed Obaiedallah - 5 years, 7 months ago

Pretty much exactly what I was thinking, too.

There's 100 possible positive integers less than 101, i.e. [1,100] is the relevant integer set to examine.

I knew intuitively that there's going to be a ton of numbers which have distinct digits, so instead I looked for the complement, which is the number which don't have two distinct digits.

11,22,33,44,55,66,77,88,99 are 9 that have repeating digits, and obviously single digit numbers would not have two distinct digits (or even two digits, at all).

Total - A^C = A Because A+A^C covers every number we're looking at ( Total ) A^C = # Integers [1,9] + integers which are factors of 11 between [1,100]. = 9 +9= 18 100 - 18 = 82

Oli Hohman - 5 years, 7 months ago

I would argue that the numbers 1-9 are 01, 02, 03, 04, 05, 06, 07, 08, 09. they contain 2 different digits, therefore, are valid. The only number that does not contain 2 different digits is 0.

Sandra Nicole - 4 years, 9 months ago

Does the number 012 contain 3 different digits then?

Leading zeroes are ignored for this reason. Let me add that to the problem.

Also, note that 0 is not a positive integer.

Chung Kevin - 4 years, 9 months ago

Thành Danh Trương
Oct 19, 2015

Jester Koh
Dec 28, 2016

The question can be reduced to how many 2-digit numbers contain two different digits since you need to have 2 digits in order for there to be a difference in the first place. But we consider 100 as well which has 2 different digits 1 and 0. Now, there are 99-10+1=90, two-digit numbers. Of which, 9 of them has the same digits, 11, 22, 33, ... , 99. So the number required is 90 - 9 +1 = 82, the addition of 1 takes into account the number 100.

Prince Loomba
Jan 22, 2016

_ _ 9 (every except 0)*9(every except utilized digit) different ways to fill the 2 places and1 for 100. Total 82.

Sadasiva Panicker
Oct 20, 2015

Multiples of 11 less than hundred = 9, one digiy number = 9, Then 100 -18 = 82.

Atika Samiha
Oct 19, 2015

2 digit number starts from10, 2 digit number less than 101 are 91 in number.but from here 9 number have same digit are. so,we can find the answar (91-9)=82

Sahil Sharma
Oct 19, 2015

Let the two digit no be 'ab'. Now to meet the given criteria a&B must be different. Thus, position of 'a' can be filled in 9 ways (1,2,3....9) & position of 'B' can also be filled in 9 ways ( excluding the digit used at 'a' but including 0). So two digit no.ab can take 9x9=81 values. However 100 also satisfies the given condition as it has 1&0 as two different digits.

Therefore total numbers satisfying above condition = 81+1=82.

Diversity of Digits

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