How many four digit numbers have no repeat digits, do not co

Question

How many four digit numbers have no repeat digits, do not contain zero, and have a sum of digits equal to 28?
A. 14
B. 24
C. 28
D. 48
E. 96

For a discussion of difficult counting questions, as well as the explanation to this question, see:
https://magoosh.com/gmat/2013/difficult- ... -problems/

Mike :-)

mau5 · Accepted Answer

Hmm. OK, lets start with the opposite of what is not allowed. We know that it's a 4 digit no and the sum of the digits has to be 28. Also, 4*7 = 28.
Let the no be 7777. However, no repetition is allowed.We try to re-arrange the given digits to form other acceptable 4 digit numbers.

Let's rearrange by reducing the value of the hundreds place by 1,and add it to the tens place : 7687. Again, take 2 units from the hundreds place and add it to the units place :  7489 . All the conditions are met. Thus, as there are 4 distinct digits, the total no of arrangements possible: 4! = 24.

Again, lets try to rearrange the given number to see if we can end up with another such unique pair of combination : Reduce the value of the thousands place by 1 and add it to the hundreds place :  6589 . Again, unique combination : 4! = 24.

Any other combination of value(s) will result in one of these 48 numbers.

Noteworthy point: The lowest of these numbers can only be of the form : 4XYZ.

farful · Answer

First, look for all 4 digits without repeat that add up to 28. To avoid repetition, start with the highest numbers first.

Start from the largest number possible 9874.
Then the next largest number possible is 9865.

After this, you'll realize no other solution. Clearly the solution needs to start with a 9 (cuz otherwise 8765 is the largest possible, but only equals 26). With a 9, you also need an 8 (cuz otherwise 9765 is the largest possible, but only equals 27). With 98__ only 74 and 65 work.

So you have two solutions. Each can be rearranged in 4!=24 ways. So 24+24=48.

Paris75 · Answer

Exactly how I did it, nothing to add!

Congrats

jlgdr · Answer

Mike, that is an awesome question One of the best I've seen from Magoooooooosh so far Cheers! J

BrainLab · Answer

checking numbers
1 x
2 x
3 x
4 ok (789)
5 ok (986)
6 ok (985)
7 ok (984)
8 ok (974)
9 ok (874)

So 6 Numbers are Ok --> 6! / (6!/4!2!) = 48

Could some expert please comment on this solution. It's not a standard way to solve such questions....

mvictor · Answer

only 2 options work:
9+8+7+4
and
9+8+6+5

total we have 4! first arrangements + 4! second arrangements
48 in total.

HardWorkBeatsAll · Answer

Let's start picking numbers from the biggest to smallest, without repetition.

Pick 9,8,7 - total is 24. To bring it to 28, let's pick 4. (Note that this also tells us that in any case, even with the largest numbers, we can't go below 4.) We have 9874. Now, we can reduce one of the bigger numbers (we'll have to reduce 7 as if we reduce 8 or 9, we'll repeat a digit) while simultaneously increasing the smallest number by 1. That way, we replace 4 with 5, and 7 with 6.
So, the combination becomes: 9865 (sum=28). These two combinations (9874 and 9865) can lead to 4! + 4! = 48 numbers.

pkm9995 · Answer

Sum can be achieved by two combinations
9874 -this can be arranged in 4! ways
9865-this can be arranged in 4! ways

so 4!+4!=48

GyMrAT · Answer

For the sum of digits to be 28, with the given constraints, we can see only two arrangements as 9847 & 9865

The # of ways to arrange the digits in each is = 4! + 4! = 48

Answer D.

Thanks,
GyM

Funsho84 · Answer

First set of numbers that satisfy is 9874. Second set is 9865.

For 9874, you can have 6 variations of the #'s per starting digit. For example:
9874
9847
9784
9748
9478
9487

Each starting digit will have 6 variations each. 6*4 = 24

The same applies to 9865

24+24= 48

Kinshook · Answer

Asked: How many four digit numbers have no repeat digits, do not contain zero, and have a sum of digits equal to 28?

{6,7,8,9} has sum of digits = 6+7+8+9 = 30
We have to reduce 2 in the digits to get sum of digits = 30 - 2 = 28
Possible combinations are = {4,7,8,9} & {5,6,8,9}
Number of 4-digits numbers formed from {4,7,8,9} = 4! = 24
Number of 4-digits numbers formed from {5,6,8,9} = 4! = 24
Total four digit numbers have no repeat digits, do not contain zero, and have a sum of digits equal to 28 = 24 + 24 = 48

IMO D

Archit3110 · Answer

total possible no ; 9874 & 9856 which can be arranged in 4!*2 ; 48 ways
IMO D

Crytiocanalyst · Answer

The posiible combinations being is 7 four times when we tru ro figure the optimal combinations we figure out 
7894=> 4! = 24 
6589=>4! = 24
IN total = 48
Hence IMO D

bumpbot · Answer

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