How many four-digit positive integers can be formed by using

[AAPL]

How many four-digit positive integers can be formed by using the digits from 1 to 9 so that two digits are equal to each other and the remaining two are also equal to each other but different from the other two?

A. 400
B. 1728
C. 108
D. 216
E. 432

The OA is D.

I am getting answer as 432, but the OA is D.

I tried to solve this PS question as follows,

The first number can be selected from 1 to 9 - in 9 ways, the second number has to be same as the one selected so only 1 way, the third number can be selected from remaining 8 numbers in 8 ways and 4th number again has to be only 1
so in total (9*1*8*1)*4!/2!*2! = 72*6 = 432.

Can anyone please explain what mistake I am making here? Thanks!

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Hi All,

We're asked for the number of different four-digit positive integers that can be formed by using the digits from 1 to 9 so that TWO digits are EQUAL to each other and the remaining two are ALSO EQUAL to each other but DIFFERENT from the other two.

The 'quirky' part about this question is that we don't know which 'pairs' of digits need to be the same - so we have to account for all possibilities. There are 3 possible ways for this result to occur (where X and Y are the digits described):

XXYY
XYXY
XYYX

XXYY: For example, 9988. Here, there are 9 options for the 1st digit and the 2nd digit has to match (so there's just 1 option), then 8 options for the 3rd digit and just 1 option for the 4th...
(9)(1)(8)(1) = 72 possible numbers

Similar patterns exist for the other two options...

XYXY: For example: 1212
(9)(8)(1)(1) = 72 possible numbers

XYYX: For example: 3443
(9)(8)(1)(1) = 72 possible numbers

Total possibilities = 72 + 72 + 72 = 216

Final Answer: D

GMAT assassins aren't born, they're made,
Rich

[Jeff@TargetTestPrep]

How many four-digit positive integers can be formed by using the digits from 1 to 9 so that two digits are equal to each other and the remaining two are also equal to each other but different from the other two?

A. 400
B. 1728
C. 108
D. 216
E. 432

There are 9C2 = 9!/(2! x 7!) = (9 x 8)/2 = 36 ways to choose 2 distinct digits from 9 digits.

Also, the number of ways to arrange the digits of a 4-digit number with two sets of digits being the same is:

number of arrangements of AABB = 4!/(2! x 2!) = 6

So the total number of ways to create the number is 36 x 6 = 216.

Answer: D