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

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!

The number of ways u can chose two digits from 9 digits is 9C2=(9*8)/2.
And the no of ways u can arrange two digits like asked is 4C2=(4*3)/2.

total ways is 36*6=216.

Hope that helps.
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Regarding the pairing you are right.It is the basic way of understanding combinations of numbers.
here you are picking 2 numbers to form a set from 9 available numbers,which turns out to be 9C2.
Basically picking of k numbers from a n set of numbers gives nCk ways of possibilities.

Hope its clear to you now.

In that solution the # of ways to choose 2 different numbers for X and Y is 9*8=72, but it should be \(C^2_9=36\), so gmihir's solution counts the same numbers twice.
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Yes, the answer should be 243.

Assume the first digit is A, for which we have 9 choices, because it cannot be 0. Then, we have 3 choices where to place the second digit A.
Now, for the other different digit B, we have again 9 choices (different from A but can be 0), and the places of the two B's are uniquely determined.
Therefore, total number of possibilities is 9*3*9 = 243.

A similar logic can be applied also for the original question:
9*3*8 = 216
9 possibilities for A
3 places to choose from for the second A
8 possibilites for B (cannot be A and cannot be 0), no places to choose, already determined
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the 4 digits can be organised in 4C2 ways = 6 ways
You have to divide by 2, so only 3 ways.
Or, 9 possibilities for the first digit, this can be placed for the second time in 3 places, then in the remaining places have another different digit - 8 possibilities.
A total of 9*3*8=216.

4C2 would work if you choose two distinct digit out of 9, which is 9C2 = 36 and you don't care about the order in which you have chosen them, take 2 of each type, then arrange them, which is 4C2 = 6.
This will give you again 9C2*4C2 = 36*6 = 216.

You cannot in one stage take order into account, then in the next stage ignore order. Choosing two digits as 9*8 means you distinguish between which one is chosen first. For example, once you considered x = 4 and y = 1, which you can arrange in 4C2 = 6 ways. But in this 6 arrangements are also include those with a 1 in the first place.
Then you considered the pair x = 1 and y = 4, for which again 4C2 = 6 arrangements, but these are identical to the previous ones. 4C2 counts arrangements without distinguishing whether you first place x or y.
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Stage 1 : Number of ways to select 2 different digits form 9 => (9 * 8 )/2 =36

Stage 2 : Number of ways to arrange AABB (MISSISSIPPI rule) = (4 * 3 *2) / (2*2) = 6

Fondamental Counting Principle = 36 * 6 = 216

Exactly, you can have 1144, or 4141, or 4411, or etc...
Since you have two pairs of equal digits, the order doesn't matter here.

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
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This is the best explanation of the conundrum by far.
I was also confused at first, not understanding why the answer is 216, not 432.

In fact, if we pick the 2 numbers as a first step using "9 x 8 = 72", we are picking for example 8 & 9 as well as 9 & 8. Which means we have already "arranged" the 2 numbers one time. And then when we arrange the 2 numbers using 4!/2!*2!, we are again arranging the same numbers, now for a second time. That is why double counting is involved.

The way to solve this is to NOT arrange the numbers when we pick them in the first place, meaning we should use "combination" instead of "permutation". So, choose 2 numbers out of 9 (9C2 = 36) and then arrange those 2 numbers in 4!/2!*2! ways.

Hope this helps