An integer is said to be “diverse” if no two of its digits are the sam

Question

An integer is said to be “diverse” if no two of its digits are the same. For example, 327 is “diverse” but 404 is not. How many “diverse” two digit numbers are there ?

A. 70
B. 72
C. 81
D. 90
E. 91

manpreetsingh86 · Accepted Answer

Let the two digit number be AB. For A we have 9 options (1,2,3,4,5,6,7,8,9), and for B we again have 9 options (0, and one of the remaining eight numbers). hence total number of ways = 9*9 = 81.

kunal555 · Answer

A two digit number has to be formed from digits 0 to 9 such that unit's digit and ten's digit are not same.
The unit's digit can have 10 values from 0 to 9.
Ten's digit can have 9 values from 1 to 9
So total number of two digit numbers will be 9 x 10 = 90
But there are 9 numbers which are not diverse.
11,22,33,44,55,66,77,88,99

so, the number of diverse two digit numbers will be 90-9 = 81

Answer:- C

Probus · Answer

Hi,

Nothing new to add but just wanted to elaborate , what kind of thinking should we use for such questions

If you look at the question very carefully its primarily asking how many two digit numbers can be made using digits from 0 to 9 when repetition is not allowed

Total available digits are 10 
So 
__ __
in first place can be filled by 9 ways ( because we cannot have a two digit number starting with 0. ). Now with one digit used up we need to fill second place. For second place we can have 0. But since we had 10 digits initially , then 1 digit used to fill first place , we are left with 9 digits . So second place can be filled by 9 ways

Now the question is Does the order matter. I mean 15 is different from 51 so yes the order matters. So its a case of Permutations. or we can say a case of AND(*) 
so we will have total of 9*9= 81 possible combinations.

APPROACH:2

we can also see it this way ( consider for time being that 0 also takes the first(ten's) place then we can fill the two spaces in 10*9= 90 ways . This 90 ways includes where 0 is in I space (ten's Place)

But in how many cases have we considered 0 in ten's place . How many ways can we fill the second (unit's) place when 0 is the only option for ten's place 9 ways

So if we subtract the cases in which we have considered 0 in ten's place from all possible cases we get the required answer. 90-9= 81

Approach:3 
Lowest two git number is 10 and highest two digit number is 99. So total number of two digit numbers are (99-10) +1 = 90
But in every set of 10 numbers beginning from 10 we will have number whose both the digits are same. And such sets are 9 so 9 such numbers between 10 & 99 where we have both digits same. 
So 90-9= 81 .

Probus

HITMAN09 · Answer

Question ask "How many “diverse” two digit numbers are there"

Total number of two digits Integer = 10 to 99 = 90

Given :- An integer is said to be “diverse” if no two of its digits are the same.

Hence, total number of diverse integer between 10 and 99 = 9 ( Multiple of 11)

Requested information = 90-9 = 81

Answer C

ScottTargetTestPrep · Answer

There are 90 two-digit numbers (from 10 to 99, inclusive). Of these integers, 9 of them (11, 22, …, 99) are not “diverse.” Therefore, there are 90 - 9 = 81 “diverse” two-digit numbers.

Answer: C

MathRevolution · Answer

The first two-digit number starts from 10 and the last two-digit number is 99.

Total numbers: 99 - 10 + 1 = 90

Non- Diverse numbers  : Both places have the same numbers: _ _

The first place will have 9 options as '0' can not be the first number. Second place will only have 1 option as it has to be the same number = 9 * 1 = 9

Diverese two digit numbers: 90 - 9 = 81

Answer C

gmatcafe · Answer

An integer is said to be “diverse” if no two of its digits are the same. For example, 327 is “diverse” but 404 is not. How many “diverse” two digit numbers are there ?

A. 70
B. 72
C. 81
D. 90
E. 91

For the first digit we have 9 choices: 1,2,3,4,5,6,7,8,9. For the units digit also we have 9 choices including 0. Total number of two digit numbers whose digits are distinct is 9*9=81 C)

mounicavenigalla · Answer

The Numbers from 90 to 100 excluding 99 and 100 we only have 8 two digits. Therefore i only see total of 8*9 = 72 + 8 =80 two digit diverse digits. can someone explain ?

Praful_0810 · Answer

We need to find two digit number.

We have 10 options (0 1 2 3 4 5 6 7 8 9)

However, since it a two digit number we can not use 0 at tens place. So we are left with 9 option.

For unit digit again we have 10 options.

Since we have constrain that any two digit can not be same we are left with 9 options.

So 9 * 9 = 81 such digits

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Sneha2021 · Answer

There are 9 diverse numbers between 90 and 99 and not 8 
#Total Numbers btw 90 and 99 including 90 & 99= 99-90+1= 10 
We need to subtract 1 number i.e 99, so diverse numbers are 10-1 = 9

Therefore total is 72+9=81.

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An integer is said to be “diverse” if no two of its digits are the sam

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