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Re: Integers less than 10,000 [#permalink]
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16 Jun 2010, 11:45
Subject: Integers less than 10,000 zestzorb wrote: The method is called "stars and bars"
I can't post the link. Look it up on google. 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?



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zestzorb wrote: The method is called "stars and bars"
I can't post the link. Look it up on google. Thanks. I looked this up and am slightly confused when to use the (n1)/(k1) vs (n+k1)/k or (n+k1)/(n1) 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. http://en.wikipedia.org/wiki/Stars_and_ ... robability) http://www.mathsisfun.com/combinatorics ... tions.html



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Re: Integers less than 10,000 [#permalink]
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20 Jul 2010, 19:06
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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Re: Integers less than 10,000 [#permalink]
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12 Oct 2010, 08:28



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Re: Integers less than 10,000 [#permalink]
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12 Oct 2010, 08:35
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



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Re: Integers less than 10,000 [#permalink]
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Re: Integers less than 10,000 [#permalink]
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14 Oct 2010, 23:03
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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Re: Integers less than 10,000 [#permalink]
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31 Jan 2011, 17:43
Bunuel wrote: Ramsay wrote: Sorry guys,
Could someone please explain the following:
"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: \(ddddd\) 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 \({n+r1}_C_{r1}\). In our case we'll get: \({n+r1}_C_{r1}={5+41}_C_{41}={8}C3=\frac{8!}{5!3!}=56\) Also see the image I found in the net about this question explaining the concept: Attachment: pTNfS2e270de4ca223ec2741fa10b386c7bfe.jpg This is a brilliant post, kudos to you Bunuel!
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Re: Integers less than 10,000 [#permalink]
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02 Jul 2011, 18:41
Terrific thread this  the explanations are very clear.
My question is  How often do we see these types of questions of the GMAT?



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Re: Integers less than 10,000 [#permalink]
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01 Oct 2011, 20:29
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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Re: Integers less than 10,000 [#permalink]
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05 Oct 2011, 02:17
Loki2612 wrote: I'm sorry but where did you see in the question that numbers like 0005 are considered 4 digits numbers? No its not a 4digit 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: Integers less than 10,000 [#permalink]
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05 Oct 2011, 08:58
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Great explanation by Bunnel.



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Re: Integers less than 10,000 [#permalink]
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02 Feb 2012, 05:42
AKProdigy87 Very simple explanation Thanks
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Re: How many positive integers less than 10,000 are there in [#permalink]
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02 Feb 2012, 21:23
Bunnel great post... thanks for the explanation



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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?
Will the stars and bars approach still work?
Thanks!



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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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How many positive integers less than 10,000 [#permalink]
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31 Oct 2012, 15:42
Hi everybody this is my first topic and i am really looking for your help This is an old Q How many positive integers less than 10,000 are there in which the sum of the digits equals 5? (A) 31 (B) 51 (C) 56 (D) 62 (E) 93 using the "stars and bars method " or "separator method". but when using the same approach with different numbers (bigger than 1o) => this method doesn't work e.g. lets find out 35 or 36 instead of 5 How many positive integers less than 10,000 are there in which the sum of the digits equals 36? the only correct answer is one integer which is 9999 => 9+9+9+9 but when using that method 39!/(36! 3!) = 9139 integers instead of 1 and the same 35 correct answer is 4 ( 9998, 9989 , 9899, 8999 ) but using the method 38!/(35! 3!) = 8436 really confusing any help



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Re: How many positive integers less than 10,000 [#permalink]
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31 Oct 2012, 21:15
ememem wrote: Hi everybody this is my first topic and i am really looking for your help This is an old Q How many positive integers less than 10,000 are there in which the sum of the digits equals 5? (A) 31 (B) 51 (C) 56 (D) 62 (E) 93 using the "stars and bars method " or "separator method". but when using the same approach with different numbers (bigger than 1o) => this method doesn't work e.g. lets find out 35 or 36 instead of 5 How many positive integers less than 10,000 are there in which the sum of the digits equals 36? the only correct answer is one integer which is 9999 => 9+9+9+9 but when using that method 39!/(36! 3!) = 9139 integers instead of 1 and the same 35 correct answer is 4 ( 9998, 9989 , 9899, 8999 ) but using the method 38!/(35! 3!) = 8436 really confusing any help The method works for 5 because each grouping of ones in this case is lesser than 10. ie, each grouping can be equated to a single digit. So the highest number that would work for this method can be only 9. Kudos Please... If my post helped.
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Re: Integers less than 10,000 [#permalink]
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25 May 2013, 22:03
Bunuel wrote: Ramsay wrote: Sorry guys,
Could someone please explain the following:
"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: \(ddddd\) 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 \({n+r1}_C_{r1}\). In our case we'll get: \({n+r1}_C_{r1}={5+41}_C_{41}={8}C3=\frac{8!}{5!3!}=56\) Also see the image I found in the net about this question explaining the concept: Attachment: pTNfS2e270de4ca223ec2741fa10b386c7bfe.jpg Hi Bunnel, Can such questions come up on gmat ?



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Re: Integers less than 10,000 [#permalink]
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26 May 2013, 03:45
Bunuel wrote: anilnandyala wrote: 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 \({n+r1}_C_{r1}\). In our case we'll get: \({n+r1}_C_{r1}={6+41}_C_{41}={9}C3=\frac{9!}{6!3!}\). Hope it's clear. Hi Bunnel, Can I say that this involves the placement of 5 identical 1's in four places such that each place can receive 0 to 5 1's.




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