In how many no. between 100 and 1000 exactly one of the digits is 3?

Question

(A) 648
(B) 512
(C) 252
(D) 225
(E) 26

Source:  https://www.GMATinsight.com

ENGRTOMBA2018 · Answer

3 scenarios possible

3XY = 1*9*9 = 81
X3Y (with X  
eq 3 or 0) = 8*1*9 = 72
Y3X (with Y  
eq 3 or 0)= 8*1*9 = 72

Total cases possible = 144+81=225

D is the correct answer.

v12345 · Answer

Another approach can be

total nos. between 100 and 1000 are 899

nos. in which all digits are 3 = 1
nos. in which digit 3 is in two places = 8 + 9 + 9 = 26 (for thousand , tens and once as different digit)
nos. in which no digit is 3 = 8*9*9 - 1 = 648 - 1 (for 100 not to be included)
required nos = 899 - 1 - 26 - 647 = 225

MathRevolution · Answer

Numbers between 100 to 1000 means from 101 to 999: Three-digit number: _ _ _.

Now as we need exactly one 3: possible combinations look like: 3 _ _, _ 3 _, _ _ 3.Overall numbers available are from 10

Case 1: 3, _,_: For a first place, we have 1 way to fix '3' and for the remaining places, we have '9' options each: 1 * 9 * 9 = 81

Case 2: _, 3,_: For a second place we have 1 way to fix '3' and for the first place, we cannot have '0' therefore we have '8' options and for third place, we have '9' options: 8 * 1 * 9 = 72

Case 3: _, _, 3: For a third-place, we have 1 way to fix '3' and for the first place we cannot have '0' therefore we have '8' options and for the second place we have '9' options: 8 * 9 * 1 = 72

Total ways: 81 + 72 + 72 = 225

Answer D

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In how many no. between 100 and 1000 exactly one of the digits is 3?

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