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

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!

This one is confusing to me. I know how to find the answer in a much more brute force manner.

Think about how many possible combinations you can make with 2 numbers (say 1 and 2)

1122, 1212, 1221, 2121, 2211, 2112 (6 combinations)

The number 1 has 8 numbers it can pair off with
The number 2 has 7 numbers it can pair off with
The number 3 has 6 numbers it can pair off with so on and so forth

\(= (8+7+6+5+4+3+2+1) * 6\)

\(= 36 * 6\)

\(= 216\)

I understood the solution of Bunuel, but can you explain where did gmihir went wrong... Thanks in advance.

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.

Hi Bunuel,

I Want to know if the statement changes from "How many four-digit positive integers can be formed by using the digits from 1 to 9" to How many four-digit positive integers can be formed by using the digits from 0 to 9", what would be answer.

As per me answer should be 243. Kindly conform it.

we units 1,2,3,4,5,6,7,8,9
total 9 digits
Now we want arragement xxyy

x - can be selected in 9 ways
x - same as first x - 1 way
y - can be selected from remaining 8 digits - 8 ways
y - same as first y - 1 way

so total# of ways = 9*1*8*1 = 72
not the 4 digits can be organised in 4C2 ways = 6 ways

So total arragements = 72*6 = 432

Am I wrong in thinking? & I didnt understand the concept mentioned by Bunuel here...

P & C is my weakest point in Quant

yeah i did but don't quite understand the logic could you pls elaborate a little?

If anyone else wanders into this forum and has the same question like Surbhi87 and I:

I though perhaps, the simple explanation for why the number of ways to pick to digits by simply multiplying 9 x 8 = 72 is because this is not a permutation, but a combination problem, where the order of selecting numbers does not matter. Am I correct?

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.

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.