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Difficulty:
55%
(hard)
Question Stats:
68% (02:33) correct
32%
(02:13)
wrong
based on 392
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How many different three-digit numbers can be formed which contain two digits that are the same, and a third digit that is different from the other two?
A. 196
B. 216
C. 243
D. 256
E. 316
A. 196
B. 216
C. 243
D. 256
E. 316
20 Mar 2023, 02:27
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How many different three-digit numbers can be formed which contain two digits that are the same, and a third digit that is different from the other two?
We have 9 choice of digits for Hundred place (as 0 can't be used), 10 choice of digits for Tens place and 10 choice of digits for Ones place.
Case I - first two digits are same and third digit is different:
9*1*9 = 81
Case II - first and third digits are same and second digit is different:
9*9*1 = 81
Case III - Last two digits are same and first digit is different:
9*9*1 = 81
So there are total of 81 + 81 + 81 = 243 such numbers.
Answer C
We have 9 choice of digits for Hundred place (as 0 can't be used), 10 choice of digits for Tens place and 10 choice of digits for Ones place.
Case I - first two digits are same and third digit is different:
9*1*9 = 81
Case II - first and third digits are same and second digit is different:
9*9*1 = 81
Case III - Last two digits are same and first digit is different:
9*9*1 = 81
So there are total of 81 + 81 + 81 = 243 such numbers.
Answer C
15 Aug 2024, 11:42
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$!vakumar.m
In Case 3, wouldnt it be 9*10*1 as 0 can be used for tens and ones place. Could you please explain? @$!vakumar.m Bunuel
