How many different 3-digit integers have exactly two different digits?

Question

(A) 1000
(B) 648
(C) 504
(D) 352
(E) 243

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Monika95 · Answer

First Case: None of the number is Zero:
Two numbers X & Y can be selected in 9*8 ways and then will be arranged in 3(factorial)/2(factorial) ways. Since, two numbers will be repeating.
So total number of ways 72*3 = 216

Second case: 0 is one of the two numbers:
Keep in mind zero can't take first place
Assume number other than zero is X
I) X is repeated twice - X will be from 1 to 9, for remaining two places zero can either take second place or third place. So total cases will be 9*2 = 18
II) Zero is repeated twice - X will be from 1 to 9, remaining two places will be filled with zero. So total cases = 9.

Total cases - 216 + 18 + 9 = 243

AnuragNair · Answer

Let us start by building cases:
Three digits : 11_ = has 9 cases (110,112,113,114,115,116,117,118,119)
 22_ = has 9 cases
Likewise till 99_ 
Total cases with first two digits same = 9*9 = 81

Now : _11 = has 8 cases (211,311,411,511,611,711,811,911)
 _22 = has 8 cases
Likewise till _99
Total cases with last two digits same = 8*9 = 72

Now : 1_1 = has 9 cases(101,121,131,141,151,161,171,181,191)
 2_2 = has 9 cases
Likewise till 9_9
Total cases with first and third digit same = 9*9 = 81

Additional cases : 100,200,300,400,500,600,700,800,900 = 9 cases

Total cases = 81 + 72 + 81 + 9 = 243.

Option E.

Sanskar115 · Answer

3-digit integers have exactly two different digits, i.e., one number same and 2 different
so, 9×1×9× \frac{ 3C2}{2! } = 81× 3 = 243

gchandana · Answer

We could do negation as well

Total - All distinct - All same (There are only three cases, all are distinct, 2 distinct and all same)

Total = 900
All distinct = 9*9*8 = 648
All same is = 9

So, 900 - 648 - 9 = 243.

Hope it helps!

pappal · Answer

need to find 3 digit integers which have exactly 2 different digits?
total 3 digit integers=900
3 digit integers with all 3 distinct digits=9*9*8=648
let (H T U): for the hundreds digit we have 9 choices from 1 to 9 , for tens place we have again 9 choices from 0 to 9 leaving the one which has occupied hundreds place and similarly for units place we have just 8 choices.
now 3 digit integers that have exactly 2 different digits and integers that have all 3 digits same=900-648=252
from here only we can see that the requisite answer will be less than 252 
so only choice that favors is  E .

How many different 3-digit integers have exactly two different digits?

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How many different 3-digit integers have exactly two different digits?

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