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22 May 2012, 09:35
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
44% (01:58) correct
56%
(02:03)
wrong
based on 840
sessions
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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
A. 400
B. 1728
C. 108
D. 216
E. 432
I am getting answer as 432, but the OA is D. Here is the approach I used.
First number can be selected from 1 to 9 - in 9 ways, 2nd number has to be same as the one selected so only 1 way, 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!
First number can be selected from 1 to 9 - in 9 ways, 2nd number has to be same as the one selected so only 1 way, 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!
22 May 2012, 10:35
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gmihir
XXYY can be arranged in \(\frac{4!}{2!2!}=6\) ways (# of arrangements of 4 letters out of which 2 X's and 2 Y's are identical);
# of ways we can select 2 distinct digits out of 9 is \(C^2_9=36\);
Total # of integers that can be formed is 6*36=216.
Answer: D.
26 Jun 2017, 17:54
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gmihir
Let’s first find the number of ways we can form a number AABB where A and B denote distinct digits.
For the first digit, we have 9 possible options. If the second digit matches that of the first, then we have 1 option. Then for digit 3, we have 8 possible options. Since digit 4 matches digit 3, we have 1 option.
So, for this scenario, we have 9 x 1 x 8 x 1 = 72 options.
Since we are essentially being asked how many ways we can arrange As and Bs in A-A-B-B, where A and B are numbers 1 through 9, we see that A-A-B-B can be arranged in 4!/(2! x 2!) = 24/4 = 6 ways.
Finally, we should take into account that we counted each number twice when we made this calculation (for instance, when A = 1 and B = 2, AABB = 1122, but when A = 2 and B = 1, BBAA = 1122); therefore, we should divide the result by 2.
Thus, the total number of ways to create a 4-digit number with 2 unique digits is (6 x 72)/2 = 216.
Answer: D
could you pls elaborate a little?
