Of the three-digit positive integers that have no digits

Question

Of the three-digit positive integers that have no digits equal
to zero, how many have two digits that are equal to each
other and the remaining digit different from the other two?
A. 24
B. 36
C. 72
D. 144
E. 216

Please explain your answers

tekno9000 · Answer

If you make a table of 3 digited nos. as follows

100-199 
in this range the three digit no with two nos. same and without two zeroes are

122, 133, 144, 155, 166, 177, 188, 199 = 8 nos.

similarly for range 200-299 = 8 nos.
300-399 = 8
400-499 = 8
500-599 = 8
.
.
.
.
900-999=8

Hence total is 72 nos.

FN · Answer

i get 216..

here is how..

1*9*8+9*8*1+8*9*1=216

vshaunak@gmail.com · Answer

Should be 216.

let's assume set1 consists of numbers having 1 as a repeatable digit. That means 11x where x is the different digit. There will be three case 11x, 1x1, x11
No digit is zero so total cases of set 1 will be 11x, 1x1, x11 = 8+8+8 =24
There will be 9 sets like 22x....33x...99x 
hence total numbers = 24*9 =216

anirudhoswal · Answer

IMO - 72

A=Total 3 digit nos with no 0 and repeating/non-repeating cases (not necessarily distinct digits) => 9*9*9
B=Total 3 digit nos with no 0 and non-repeating cases only (distinct digits only) => 9*8*7
C=Total 3 digit nos with no 0 and all 3 digits same => 9 (111, 222, 333, ... )

We need 2 nos repeating, i.e. , A-B-C => 9 (81-72) - 9 => 72

goalsnr · Answer

Thanks for sharing the systematic approach. The OA is E

anirudhoswal · Answer

Correction in my last statement above,
We need 2 nos repeating, i.e. , A-B-C => 9 (81- 56 ) - 9 =>  216

Of the three-digit positive integers that have no digits

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