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

Wouldn't the 4!/2!×2! Also select xyxy which is not to be selected?
Bunuel?

Posted from my mobile device

The 4-digit number comes in 3 ways: AABB, ABBA, ABAB

For AABB:
Number of options for the first "A" is 9, for the second "A" is 1 since the second "A" has to be the same as the first "A"
Number of options for the first "B" is 8, since it can't be the same as "A." One option for the second "B" since it has to be the same as the first "B"
Therefore, number of options for selecting a "AABB" style 4-digit number is 9*8=72

The same goes for ABBA and ABAB.

Therefore, number of options we have for the 4-digit number is 72*3=216.

to answer the OP as shortly as possible:


you are "Double Arranging" when you set it up that way. I followed your same path, but then realized when you are arranging the 9 different options in the 1 slot and then 8 different options in the other slot, you have already arranged these when you performed the 4!/2!*2! (presumably to find how many ways we can arrange: aabb ---Type of Number)

1st) Number of ways we can arrange the Number of Type: aabb --- in which 2 Digits are the same and a Different 2 Digits are the Same

4! / (2!*2!) = 6 Scenarios


AND

2nd) for Any 1 Arrangement, the Different Combinations of Numbers are the Following:

Assuming Scenario: aa bb

9 Options ----- Locked 1 Option ------ 8 Options Remaining ------ Locked 1 Option

9 * 1 * 8 *1 / 2! ----- Divide out by 2! to avoid over-counting because we do NOT want to count the Arrangements in which the a and b switch. We already did that in Step 1.


Answer:

(4! / 2!*2!) * (9 * 1 * 8 * 1 / 2!) =

6 * (72/2) =

6 * 36 = 216 Numbers that follow the constraints.



The Other Method is to 1st Choose which 2 Distinct Digits are going to be the Repeats that show up in the 4 Digit Number

1st) Selection/Choosing

You can Choose which 2 Distinct Digits that will Repeat in the Number in the following ways:

"9 choose 2" Ways = 36 ways

AND

2nd)ARRANGEMENT

For Each Combination of 2 Distinct Digits Chosen, We can Arrange them in the following ways:

4! / 2! * 2! ------ because 2 of the Elements will be Indistinguishable Elements (2a 2b 3 3) is the same Arrangment as (2b 2a 3 3)


36 * 6 = 216 Ways

-D-

How come the approach by gmihir doesn't result in the same answer? He reduced the possibilities for the 2nd digit to 8.

How come this question is different from this one: https://gmatclub.com/forum/of-the-three ... 41663.html

I used the same approach here: (9*1*8*1)*4!/2!*2! = 72*6 = 432, but got it wrong

The 4-digit number comes in 3 ways: AABB, ABBA, ABAB why only this
why not BBAA and BABA too, are they not diff arrangements.

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.