Problem Solving

lol that really rude guy got it wrong

hiojay, could you explain how you knew to do 9 x 9 x 1 x 1 etc? I faintly remember combinations and permutations from IB Math Studies! So there are 6 possible combinations...that's as far as I got

4C2 means that out of 4 possible positions, we want to find out how many ways we can select groups of 2 (order doesn't matter).

In the context of this question, we basically want to find out how many ways we can select groups of 2 digits (order doesn't matter) from a single group of 4 digits.

The formula for combinations is:
n! / k!(n - k)!

When we substitute the values for n and k in the formula we get:
4! / 2!(4 - 2)!

This can be simplified to (4 x 3) / (2 x 1) = 6

Now we need to visualise how those combinations can be arranged:
1. 33XX
2. 3X3X
3. 3XX3
4. X33X
5. X3X3
6. XX33

In 1, 2 and 3: X can be any value from 0 to 9, except for 3.
In 4, 5 and 6: Take note that the X value in the first digit cannot be 0 or 3. The other X value can be any value from 0 to 9, except for 3.

So for the first combination 33XX:
We can see that there is only 1 possibility in the first and second digits (i.e. a 3 in both), and there are 9 possibilities in the 3rd and 4th digits (i.e. any number from 0-9 excluding 3).

When we multiply this out we get 1 x 1 x 9 x 9 = 81.

If you're confused about multiplying it out, then let's digress for a moment.

--- START OF DIGRESSION ---

If you had 2 six-sided die and rolled them both at the same time, then how many possible outcomes could you have?

We know that there are 6 outcomes for the first dice (1,2,3,4,5,6), and then for every outcome of the first dice, we know that there are another 6 outcomes for the second dice (eg. [1,1], [1,2], [1,3], [1,4], [1,5], [1,6], [2,1],[2,2], etc.).

That gives us 6 x 6 = 36 possible outcomes.

--- END OF DIGRESSION ---

So going back to the original question, we can see that for combination 1 we can have only 1 possibility in the first and second positions, and 9 possibilities in the third and fourth positions.

That means you can have 81 different 4-digit numbers that have 33 as the first two digits, and don't have any remaining 3s in the last two digits.

If we follow this formula for all of the combinations we get:
1. 1 x 1 x 9 x 9 = 81
2. 1 x 9 x 1 x 9 = 81
3. 1 x 9 x 9 x 1 = 81
4. 8 x 1 x 1 x 9 = 72
5. 8 x 1 x 9 x 1 = 72
6. 8 x 9 x 1 x 1 = 72

4, 5, and 6 all have 8 ways for the first digit since 0 and 3 are invalid values.

Add all the possibilities together and you get 459.

Hope that is clear.

Thanks! That totally makes sense! I remember getting a question like this before, and I basically sat there and wrote out all the possible options.

OOH maybe you know how to do this question too:

How many three digit numbers are there that are over 700 and have two of the same digit with the remaining digit a different number?

[spoiler]1.) 77x = 1x1x9 = 9
2.) 7x7 = 1x9x1 = 9
3.) 88x = 1x1x9 = 9
4.) 8x8 = 1x9x1 = 9
5.) 99x = 1x1x9 = 9
6.) 9x9 = 1x9x1 = 9
7.) x00 = 2x1x1 = 2
8.) x11 = 3x1x1= 3
9.) x22 = 3x1x1= 3
10.) x33 = 3x1x1= 3
11.) x44 = 3x1x1= 3
12.) x55 = 3x1x1= 3
13.) x66 = 3x1x1= 3
14.) x77 = 2x1x1= 2
15.) x88 = 2x1x1= 2
16.) x99 = 2x1x1= 2

Total: 80[/spoiler]

Is that correct?

gmat740 wrote:@quriousaddict

Well if in any way I sound rude to you, then I really apologize for that.
You are replies and attempt to solve the questions will be very much welcome to this forum like others.

But,

Can you have the number that starts with 0xxx or 0033 as a 4-digit number?

I won't support you on this at least.Being new to the forum is an alibi for your silly mistake.

You can ask a 6 year old neighborhood kid about the question you posted,may be he has a better answer to it!!

Being new to the forum does not mean you lack common sense, don't you think?

We are all humans and by nature we commit mistakes. I would advice you to approach every question you face in a logical way without assuming anything from your side(trust me assuming anything from your side is really fatal in Verbal part!!)
I would appreciate to see more answers from you in the future.
In case of any advice don't hesitate to pm me.

Cheers,

Karan

Karan,
Any technical explanation whatever good pr bad it is doesn't demands the language that you have used, there are many things that you post that is illogical to me in even wider sense, that doesnt mean I start bursting on you, why you have posted such silly stuff,

You must apolize to the poster here unconditionally, in my opinion.

Yes mixpanda, that looks correct to me. Good job!

hiojay wrote:I'm counting 459 numbers too.

I see 4C2 or 6 possible combinations that we need to address.

1. 33XX
2. 3X3X
3. 3XX3
4. X33X
5. X3X3
6. XX33

Looking at the number of possibilities for each combination:
1. 1 x 1 x 9 x 9 = 81
2. 1 x 9 x 1 x 9 = 81
3. 1 x 9 x 9 x 1 = 81
4. 8 x 1 x 1 x 9 = 72
5. 8 x 1 x 9 x 1 = 72
6. 8 x 9 x 1 x 1 = 72

4, 5, and 6 all have 8 ways for the first digit since 0 and 3 are invalid values.

Add all the possibilities together and you get 459.

thanks man, this is a really neat explanation. im amazed.

An essentially similar way is to consider all the possible 4-digit combinations then to subtract those which begin with zero.

Total: 4C2 x (1C1)^2 x (9C1)^2 = 6x1x1x9x9 = 486

Beginning with 0: 1C1 x 3C2 x (1C1)^2 x 9C1 = 1x3x1x1x9 = 27

Ans: 486-27 = 459