arjunshn wrote:How many 4 digit integers can be formed from 1-9 numbers such that two digits have the same value and the other two have the same value but different from the first two?

ans-216

can someone explain hw..i am getting 432= 72*4!/2!*2!

I always suggest that students start listing possible outcomes (either in your head or on paper). Doing so may provide some insight into the solution. So here, we're looking for 4-digit numbers with 2 pairs of digits. Some outcomes include 2255, 6776, 8181 etc.

Here, I can see that to list the outcomes, first I need to choose the 2 digits to work with.

Then, once I select my 2 digits, I need to arrange two of 1 digit and two of the other digit. For example, let's say I choose the two digits 3 and 7. This means I need to arrange two 3's and two 7's.

As we can see, we can break the task of listing possible outcomes and break it into 2 stages:
Stage 1: Choose the 2 digits to work with
Stage 2: Arrange the 4 subsequent digits

Stage 1: There are 9 digits to choose from and, for this first step, the order of the selected digits does not matter. So, this is a combination question.
We can select 2 digits from 9 digits in 9C2 ways.
This is equal to 36

Stage 2: We want to arrange 4 digits, where there are 2 pair of each.
One option is to just list the possibilities. For example, say the two digits are 3 and 7.
We can arrange two 3's and two 7's in the following ways: 3377, 3737, 3773, 7733, 7373, 7337 (6 ways).
Another way to determine this is to treat the arrangement as a MISSISSIPPI question, where we are arranging objects where some of the objects are identical.
Here we have 4 objects and we have 2 identical digits, and another 2 identical digits. So, we can arrange these 4 objects is 4!/2!2! ways. This is equal to 6

From here we can apply the Fundamental Counting Question and multiply the results from each stage to get 36 x 6 = 216

For more information on the Fundamental Counting Question, watch video #3 at https://www.gmatprepnow.com/module/counting

force5 wrote:Brent can i ask you a question why are we not apply fundamental principle of counting 9*8 to choose two numbers. more over our number is in the form of aabb. you are selecting a and b by doing 9C2 why are you not arranging them further by 2C1 ways. because the order matters here.
i face a lot of difficulty in assessing this. Is there a way to understand when order matters and when it doesnt.

Hi Force5,

Good question. The question "does order matter?" is not always easy to answer.
In my solution, I broke the task of "building" outcomes and broke it into 2 stages.
Stage 1 was just selecting 2 digits to work with. At that point, I wasn't arranging anything. So, selecting the digits 3 and 7 is exactly the same as selecting the digits 7 and 3.
Since the order of the two selected digits does not matter during this stage, this particular stage (stage 1) represents a combination question.
So, as far as stage 1 is concerned, we can selected 2 digits from 9 digits in 9C2 ways. (by the way, I have a free video for quickly calculating combinations in your head at https://www.gmatprepnow.com/module/counting - it's video #17)
9C2 = 36, so there are 36 different ways in which we can select 2 digits to work with.

In stage 2, we are essentially taking each of these 36 different pairs of digits and asking, "In how many ways can I arrange the 4 subsequent digits?"
Well, for the digits 3 and 7, we saw that the 4 subsequent digits can be arranged in 6 ways.
By the same token, the two 8's and two 5's, can be arranged in 6 ways.
And two 1's and two 9's, can be arranged in 6 ways.

Since each of the 36 pairs of digits can be arranged in 6 ways, the total number of outcomes is 36 x 6 = 216

We want to count how many 4-digit numbers look like 1188, or 4545, or 7227. Another way to look at the problem:

* for the thousands digit, I have 9 choices
* once I've chosen the thousands digit, one of the 3 remaining digits must be the same as the first. I have 3 choices for which digit will be equal to the thousands digit (it can be the hundreds, tens, or units digit)
* finally the remaining two digits must be equal to each other and different from the thousands digit, so I have 8 choices for that number

Multiplying our choices gives the answer: 9*3*8 = 216.