_ _ _ _

First slot 9 ways (excluding 0)
Second 9 ways
Third 8 ways

9*9*8*3 (Last can be filled only in 3 ways as one of the previous 3 integers should be used.

IMO A

The answer is A

Let the Four digit number be ABCD
Now, A can be filled in 9 ways (removing 0. If 0 is in A position then it will be a 3 digit number )

B can be filled in 9 ways again (0-9 excluding the number used in A)
similarly C can be filled in 8 ways (0-9 excluding the numbers used in A & B)
Now D can be filled in 3 ways (as the 4 digit number should have exactly 3 distinct digits, D has be one of the numbers used in A,B,C)

so the total number of 4 digit numbers formed = 9*9*8*3 = 1944

Regards,
Chetan

knight247 wrote:How many 4 digit numbers can be formed by using the digits 0-9 so that it contains exactly 3 distinct digits?
(A)1944
(B)3240
(C)3850
(D)3888
(E)4216

Don't have an OA. Detailed explanations would be appreciated

Case 1: Tens digit and units digit the same:
Number of options for the thousands digit = 9. (Any digit 1-9)
Number of options for the hundreds digit = 9. (Any digit 0-9 not yet chosen)
Number of options for the tens digit = 8. (Any digit 0-9 not yet chosen)
Number of options for the units digit = 1. (Must be the same as the tens digit)
To combine the options above, we multiply:
9*9*8*1 = 648.

Other cases:
Of the 4 digits, any pair could be the two that are the same.
Number of combinations of 2 that can be formed from 4 options = 4C2 = 6.

To combine the options above, we multiply:
648*6 = 3888.

The correct answer is D.

GMATGuruNY wrote:

knight247 wrote:How many 4 digit numbers can be formed by using the digits 0-9 so that it contains exactly 3 distinct digits?
(A)1944
(B)3240
(C)3850
(D)3888
(E)4216

Don't have an OA. Detailed explanations would be appreciated

Tens digit and units digit the same:
Number of options for the thousands digit = 9. (Any digit 1-9)
Number of options for the hundreds digit = 9. (Any digit 0-9 not yet chosen)
Number of options for the tens digit = 8. (Any digit 0-9 not yet chosen)
Number of options for the units digit = 1. (Must be the same as the tens digit)
To combine the options above, we multiply:
9*9*8*1 = 648.

Other options:
2 of the 4 digits must be the same.
Number of combinations of 2 that can be formed from 4 options = 4C2 = 6.

To combine the options above, we multiply:
648*6 = 3888.

The correct answer is D.

I could not understand the step of Other options (highlighted above).
Isn't it the units digit has to one among other 3 digits to satisfy the condition of 3 digits being distinct? :roll:

Please correct me where did I go wrong

Regards,
Chetan

Same doubt as Chetan, 4C2 is used when you have 4 different Nos, you got only 3 different Numbers, why would you use 4C2?

chetansharma wrote: Tens digit and units digit the same:
Number of options for the thousands digit = 9. (Any digit 1-9)
Number of options for the hundreds digit = 9. (Any digit 0-9 not yet chosen)
Number of options for the tens digit = 8. (Any digit 0-9 not yet chosen)
Number of options for the units digit = 1. (Must be the same as the tens digit)
To combine the options above, we multiply:
9*9*8*1 = 648.

Other options:
2 of the 4 digits must be the same.
Number of combinations of 2 that can be formed from 4 options = 4C2 = 6.

To combine the options above, we multiply:
648*6 = 3888.

The correct answer is D.

I could not understand the step of Other options (highlighted above).
Isn't it the units digit has to one among other 3 digits to satisfy the condition of 3 digits being distinct? :roll:

Please correct me where did I go wrong

Regards,
Chetan[/quote]

3 distinct digits means that 2 positions in the number must contain the same digit, with the remaining 2 positions containing different digits.

Case 1:
9*9*8*1 = 648 is the number of possible integers if the TENS DIGIT and the UNITS DIGIT are the same:

Now we need to count the other PAIRS of digits that could be the same:
The number of possible integers if the HUNDREDS digit and the UNITS digit are the same.
The number of possible integers if the THOUSANDS digit and the UNITS digit are the same.
The number of possible integers if the HUNDREDS digit and the TENS digit are the same.
The number of possible integers if the THOUSANDS digit and the TENS digit are the same.
The number of possible integers if the THOUSANDS digit and the HUNDREDS digit are the same.

Of the 4 digits, ANY PAIR could contain the duplicate integers.
Including Case 1, there are 6 possible pairs.
Thus, we multiply by 6:
6*648 = 3888.

GMATGuruNY wrote:3 distinct digits means that 2 positions in the number must contain the same digit, with the remaining 2 positions containing different digits.

Case 1:
9*9*8*1 = 648 is the number of possible integers if the TENS DIGIT and the UNITS DIGIT are the same:

Now we need to count the other PAIRS of digits that could be the same:
The number of possible integers if the HUNDREDS digit and the UNITS digit are the same.
The number of possible integers if the THOUSANDS digit and the UNITS digit are the same.
The number of possible integers if the HUNDREDS digit and the TENS digit are the same.
The number of possible integers if the THOUSANDS digit and the TENS digit are the same.
The number of possible integers if the THOUSANDS digit and the HUNDREDS digit are the same.

Of the 4 digits, ANY PAIR could contain the duplicate integers.
Including Case 1, there are 6 possible pairs.
Thus, we multiply by 6:
6*648 = 3888.

I guess both shankar and myself were thinking on the same lines.
It never crossed my mind that even other digits can be same not just units digit with other three.

Regards,
Chetan
Thanks for the explanation

Here's an alternate approach.

Which 2 positions in the 4-digit integer can contain the same digit?
Number of combinations of 2 that can be formed from 4 choices = 4C2 = 6.

If we disregard that the thousands digit cannot be 0, how many options for this pair?
10. (Any digit 0-9.)

How many options for the 3rd digit?
9. (Any digit 0-9 other than the digit used in the pair above.)

How many options for 4th digit?
8. (Any digit 0-9 other than the 2 digits already used.)

Multiplying the options above:
6*10*9*8 = 4320.

A fraction of these 4320 options are not allowed because they put 0 in the thousands place.
Since each of the 10 digits has the same probability of appearing in the thousands place, 1/10 of the 4320 options will put 0 in the thousands place.

Thus, 9/10 of the 4320 options are viable:
(9/10)*4320 = 3888.

GMATGuruNY wrote:Here's an alternate approach.

Which 2 positions in the 4-digit integer can contain the same digit?
Number of combinations of 2 that can be formed from 4 choices = 4C2 = 6.

If we disregard that the thousands digit cannot be 0, how many options for this pair?
10. (Any digit 0-9.)

How many options for the 3rd digit?
9. (Any digit 0-9 other than the digit used in the pair above.)

How many options for 4th digit?
8. (Any digit 0-9 other than the 2 digits already used.)

Multiplying the options above:
6*10*9*8 = 4320.

A fraction of these 4320 options are not allowed because they put 0 in the thousands place.
Since each of the 10 digits has the same probability of appearing in the thousands place, 1/10 of the 4320 options will put 0 in the thousands place.

Thus, 9/10 of the 4320 options are viable:
(9/10)*4320 = 3888.

Awesome.
Loved the alternate approach

tabsang wrote:

GMATGuruNY wrote:Here's an alternate approach.

Which 2 positions in the 4-digit integer can contain the same digit?
Number of combinations of 2 that can be formed from 4 choices = 4C2 = 6.

If we disregard that the thousands digit cannot be 0, how many options for this pair?
10. (Any digit 0-9.)

How many options for the 3rd digit?
9. (Any digit 0-9 other than the digit used in the pair above.)

How many options for 4th digit?
8. (Any digit 0-9 other than the 2 digits already used.)

Multiplying the options above:
6*10*9*8 = 4320.

A fraction of these 4320 options are not allowed because they put 0 in the thousands place.
Since each of the 10 digits has the same probability of appearing in the thousands place, 1/10 of the 4320 options will put 0 in the thousands place.

Thus, 9/10 of the 4320 options are viable:
(9/10)*4320 = 3888.

Awesome.
Loved the alternate approach

Hey Mitch,

I went over the solution again and I have a doubt.

I get the 4C2=6 part.

I was wondering if disregarding that 0 cannot be in the thousand's place is necessary.
Can't we still arrive at the solution even if we considered that 0 can't be in the thousand's place?

That way, we'll be saved the 1/10, 9/10 trouble.

For eg:
***Copy-paste from your prev. response, albeit with modifications***

Which 2 positions in the 4-digit integer can contain the same digit?
Number of combinations of 2 that can be formed from 4 choices = 4C2 = 6.

How many options for the thousand's place?
9. (Any digit 1-9.)

How many options for the 2nd (or 3rd) digit?
9. (Any digit 0-9 other than the digit used in the pair above.)

How many options for 3rd (or 4th) digit?
8. (Any digit 0-9 other than the 2 digits already used.)

Multiplying the options above:
6*9*9*8 = 3888.

I tried applying the same logic for a 5 digit number with 4 distinct digits:
Which 2 positions in the 5-digit integer can contain the same digit?
Number of combinations of 2 that can be formed from 5 choices = 5C2 = 10.

And, using the same strategy as the one used for the 4-digit problem, we get
10*9*9*8*7

Do let me know your thoughts.

Cheers,
Taz