manpreetsingh86 wrote:
manhattan187 wrote:
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
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.
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:3Lowest 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
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Probus
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