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Start the planHow 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!
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!
