How many different three-digit numbers can be formed which contain two

Question

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

$!vakumar.m · Answer

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

Jaychoudhary · Answer

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

Bunuel · Answer

No, 9*9 is correct. 9 options for the first digit and 9 options for the remaining digits (any digit except the one used for the first digit).

N0BU · Answer

So this is how I broke it down, hope it helps.

DanTheGMATMan · Answer

Classic permutation, leftmost digit can't be zero, so 9 options, then an identical digit (so 1 options) and then 9 remaining options for the third digit, and remember to multiply by how many different orders you could have drawn this in:

https://www.youtube.com/watch?v=vkVwLRfskmE

ravi4470 · Answer

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

HridhayKapoor · Answer

because if we take the second and third digits as 0 then we have 9 options for the first digit but, if we take it as any digit from 1-9 then the first digit has 8 choices - 9 x 1 + 9 x 8 = 9+72 = 81.

better to start with the most restrictive digit i guess which more often than not includes the first one.

even im also learning this haha but i understood it this way, hope it helps !

Adit_ · Answer

Choosing 3 digits:
10C1*1*9C1 = 90

Arranging them:
90*3 = 270.

We need to subtract cases with 0 as first digit:
Case 1:
When other digit is also 0:
9*2 ways = 18.

Case 2:
When the other two digits are same other than 0:
9 ways.

Total = 270-18-9 =  243 .

Answer:  Option C

____________________________

Honestly feel 700+ level because of the trap to subtract 2 cases with 0 at the start and one can't do the usual 9*8 or 9*9 directly to subtract bad cases as usually happens in such questions

How many different three-digit numbers can be formed which contain two

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How many different three-digit numbers can be formed which contain two

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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