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Basically, the question asks how many 4 digit numbers (including those in the form 0XXX, 00XX, and 000X) have digits which add up to 5. Think about the question this way: we know that there is a total of 5 to be spread among the 4 digits, we just have to determine the number of ways it can be spread.

Let X represent a sum of 1, and | represent a seperator between two digits. As a result, we will have 5 X's (digits add up to the 5), and 3 |'s (3 digit seperators).

So, for example:

XX|X|X|X = 2111 ||XXX|XX = 0032

etc.

There are 8C3 ways to determine where to place the separators. Hence, the answer is 8C3 = 56.

"There are 8C3 ways to determine where to place the separators"

I'm not familiar with this shortcut/approach.

Ta

Consider this: we have 5 \(d\)'s and 3 separators \(|\), like: \(ddddd|||\). How many permutations (arrangements) of these symbols are possible? Total of 8 symbols (5+3=8), out of which 5 \(d\)'s and 3 \(|\)'s are identical, so \(\frac{8!}{5!3!}=56\).

With these permutations we'll get combinations like: \(|dd|d|dd\) this would be 3 digit number 212 OR \(|||ddddd\) this would be single digit number 5 (smallest number less than 10,000 in which sum of digits equals 5) OR \(ddddd|||\) this would be 4 digit number 5,000 (largest number less than 10,000 in which sum of digits equals 5)...

Basically this arrangements will give us all numbers less than 10,000 in which sum of the digits (sum of 5 d's=5) equals 5.

Hence the answer is \(\frac{8!}{5!3!}=56\).

Answer: C (56).

This can be done with direct formula as well:

The total number of ways of dividing n identical items (5 d's in our case) among r persons or objects (4 digt places in our case), each one of whom, can receive 0, 1, 2 or more items (from zero to 5 in our case) is \({n+r-1}_C_{r-1}\).

In our case we'll get: \({n+r-1}_C_{r-1}={5+4-1}_C_{4-1}={8}C3=\frac{8!}{5!3!}=56\)

Also see the image I found in the net about this question explaining the concept:

Attachment:

pTNfS-2e270de4ca223ec2741fa10b386c7bfe.jpg [ 63.83 KiB | Viewed 52083 times ]

Bunnel I have solved it in this way but your methods seems to be quicker. But I couldn't understand. Could you please explain in simple words.
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Basically, the question asks how many 4 digit numbers (including those in the form 0XXX, 00XX, and 000X) have digits which add up to 5. Think about the question this way: we know that there is a total of 5 to be spread among the 4 digits, we just have to determine the number of ways it can be spread.

Let X represent a sum of 1, and | represent a seperator between two digits. As a result, we will have 5 X's (digits add up to the 5), and 3 |'s (3 digit seperators).

So, for example:

XX|X|X|X = 2111 ||XXX|XX = 0032

etc.

There are 8C3 ways to determine where to place the separators. Hence, the answer is 8C3 = 56.

This is correct. Also this is the best way to solve this question. +1. (Solved exactly the same way)

Bunnel I have solved it in this way but your methods seems to be quicker. But I couldn't understand. Could you please explain in simple words.

this approach was more natural and my first idea to solve the promblem

Then I saw the Bunnel´s and AKProdigy87´s way to solve this problem.

The idea of Bunnel and AKProdigy87 method is that the sum of the digits must equal 5, and this five can be distributed among the 4 digits and numbers are made by "ones". Again, numbers are made by "ones"! please do not thing of numbers as 2, 3, 4 or 5 digit for example:

thanks Bunuel can u explain me this by using the formulae How many positive integers less than 10,000 are there in which the sum of the digits equals 6? thanks in advance

6 * (digits) and 3 ||| --> ******||| --> # of permutations of these symbols is \(\frac{9!}{6!3!}\).

Or: The total number of ways of dividing n identical items (6 *'s in our case) among r persons or objects (4 digt places in our case), each one of whom, can receive 0, 1, 2 or more items (from zero to 6 in our case) is \({n+r-1}_C_{r-1}\).

In our case we'll get: \({n+r-1}_C_{r-1}={6+4-1}_C_{4-1}={9}C3=\frac{9!}{6!3!}\).

Can this method be used in variations of this question?

Such as sum of 7 digits that add to equal 6?

Yes. Why not? You mean numbers with 7 digits? I don't know if that is going to be a question of gmat, but... If the question was to equal 5 you can do the same to six, just add one X

I am kinds of new to this forum and i am excited to see the happenings here. you guys are simply awesome . I have a question, will we get these kinds of complicated questions in GMAT?

Thanks. I looked this up and am slightly confused when to use the (n-1)/(k-1) vs (n+k-1)/k or (n+k-1)/(n-1)

the / does not indicate division. Check this out link and then the link below (scroll to bottom of page for the second link). The second link uses theorem 2. Not sure I really understand the difference between the two.

So let's say we have the same question but we want to sum to 6 instead of 5 - then we use \(C^9_3\) = 84. If we wanted to find numbers below 100,000 that the digits sum to 5 we would use \(C^9_4\) = 126.

Really elegant guys. Thanks.
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