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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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If you find my posts useful, please award me some Kudos!

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

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: \({n+r-1}_C_{r-1}={6+4-1}_C_{4-1}={9}C3=\frac{9!}{6!3!}\).

Man this is one of the hardest questions I have seen. Couldn't really understand at first (still not a firm grasp) and couldn't solve. After following the above posts it is coming to mind slowly but I am gonna go find more about "Stars & Bars" on internet (Looks like need tutor here ).
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"I choose to rise after every fall" Target=770 http://challengemba.blogspot.com Kudos??

"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

This is a brilliant post, kudos to you Bunuel!
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Thank you for your kudoses Everyone!!!

"It always seems impossible until its done." -Nelson Mandela

Hi want to know hy two digit Numbers are neglected when forming combinations can anyone explain me in a breif manner i am totally confused.
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I'm sorry but where did you see in the question that numbers like 0005 are considered 4 digits numbers?

No its not a 4-digit number.

Question says, "integers less than 10000". That includes: 0005->Single digit 0050->2 digits 0500->3 digits 5000->4 digits

The idea is that "5" must not be ignored while counting. The method suggested ensures that 5 is not ignored and is counted only once. The representation may very well be "0005", yet it is still a single digit 5.
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Re: How many positive integers less than 10,000 are there in [#permalink]

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07 Feb 2012, 15:23

I understand that the total number of positive integers less than 10000 with the sum of the digits equal to 5 is 8C3 = 56.

Any idea how to solve this if the question instead is to find the total number of 4 digit numbers less than 10000 with the sum of digits equal to 5. Is it 8C2 = 28?

Re: How many positive integers less than 10,000 are there in [#permalink]

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23 May 2012, 16:44

Bunuel,

Many apologies for reviving this method. But I gotta say I love your "stars and bars" method. My Question is it seems to only work for this question if the sum is less than 9 for this question. You can't have more than 9 stars in each slot. What if the question had asked,

How many positive integers less than 10,000 are there in which the sum of the digits equals 13?

Now this is a little tricker. Here is another example from another problem posted on this forum that illustrates the problem. Is there a way we can exand "stars and bars" to numbers that sum larger than 9?

A wheel of fortune contains numerical values from 1 to 8. The scoring system of the game is based on the sum of these values. If the host were to spin the wheel three times, how many possible number of combinations are there that will give the player the sum of 16 points?

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