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Magoosh GMAT Instructor
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Re: How many four digit numbers have no repeat digits, do not co [#permalink]
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farful wrote:
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.


Exactly how I did it, nothing to add!

Congrats
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Re: How many four digit numbers have no repeat digits, do not co [#permalink]
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mikemcgarry wrote:
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 :-)


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

Cheers!
J :)
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How many four digit numbers have no repeat digits, do not co [#permalink]
2
Kudos
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....
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How many four digit numbers have no repeat digits, do not co [#permalink]
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.
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Re: How many four digit numbers have no repeat digits, do not co [#permalink]
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.
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Re: How many four digit numbers have no repeat digits, do not co [#permalink]
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
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Re: How many four digit numbers have no repeat digits, do not co [#permalink]
mikemcgarry wrote:
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 :-)



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
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Re: How many four digit numbers have no repeat digits, do not co [#permalink]
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
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Re: How many four digit numbers have no repeat digits, do not co [#permalink]
mikemcgarry wrote:
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 :-)


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
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Re: How many four digit numbers have no repeat digits, do not co [#permalink]
mikemcgarry wrote:
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 :-)


total possible no ; 9874 & 9856 which can be arranged in 4!*2 ; 48 ways
IMO D
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Re: How many four digit numbers have no repeat digits, do not co [#permalink]
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
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Re: How many four digit numbers have no repeat digits, do not co [#permalink]
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Re: How many four digit numbers have no repeat digits, do not co [#permalink]
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