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How many positive integers less than 10,000 are there in
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Updated on: 13 Oct 2009, 21:27
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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
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Originally posted by Bunuel on 13 Oct 2009, 20:37.
Last edited by Bunuel on 13 Oct 2009, 21:27, edited 1 time in total.




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Re: Integers less than 10,000
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07 Apr 2010, 03:50
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 [ 63.83 KiB  Viewed 81160 times ]
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Re: Integers less than 10,000
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14 Oct 2009, 12:23
I believe the answer to be C: 56.
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:
XXXXX = 2111 XXXXX = 0032
etc.
There are 8C3 ways to determine where to place the separators. Hence, the answer is 8C3 = 56.




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Re: Integers less than 10,000
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14 Oct 2009, 20:06
AKProdigy87 wrote: I believe the answer to be C: 56.
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:
XXXXX = 2111 XXXXX = 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) Answer: 56.
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Re: Tough Combinatronics questions
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11 Jan 2010, 13:15
Since the sum of the 4 digits have to equal to 5, anyting #s >=6000 is eliminated.
5000s range  1 5000 is the only possibly number in the 5000's range.
4000s range  3 Since 4+1=5, there are only 3 spots to put the 1s, namely 4100, 4010, 4001.
3000s range  6 We can have 3+2+0+0 (3 combinations) or 3+1+1+0 (3 combinations).
2000s range  7 We can have 2+2+1+0 (6 combinations because of 3! ways of arranging the last 3 digits using 2,1,0) or 2+1+1+1 (1 combination).
1000s range  15 We can have 1+1+1+2 (3 combinations), 1+2+2+0 (3 combinations), 1+4+0+0 (3 combinations), and 1+3+1+0 (6 combinations because of 3! ways of arranging the last 3 digits of 3,1,0).
The # of numbers is therefore 1+3+6+7+15=32. Notice that you get 3 combinations whenever 2 of the last 3 digits are the same.
Then use the same logic on 3 digits and 2 digits and 1 digit numbers.
Is there a fast way to do this?



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Re: Tough Combinatronics questions
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11 Jan 2010, 16:46
zaarathelab wrote: 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 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: \({n+r1}_C_{r1}={5+41}_C_{41}={8}C3=\frac{8!}{5!3!}=56\)
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Re: Tough Combinatronics questions
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11 Jan 2010, 18:17
What's the concept behind the separators?



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Re: Tough Combinatronics questions
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11 Jan 2010, 23:58
mrblack wrote: What's the concept behind the separators? This is what I found in the net about this question explaining the concept: Attachment:
pTNfS2e270de4ca223ec2741fa10b386c7bfe.jpg [ 63.83 KiB  Viewed 26792 times ]
Hope it's clear.
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Re: Integers
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10 Mar 2010, 23:31
there is a shortcut. For the problem, 4 digits are equally important in 00009999 set and it is impossible to build a number using only one digit (like 11111) So, answer has to be divisible by 4. Only 56 works. Posted from GMAT ToolKit
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Re: Integers less than 10,000
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13 May 2010, 16:08
Pairs possible : 1,2,1,1 ; 1,3,1,0 ; 1,4,0,0 ; 0,1,2,2 ; 0,0,0,5 ; 0,0,2,3 which gives : 1,2,1,1 => 4!/3! = 4 0,0,0,5 =>=> 4!/3! = 4 1,3,1,0 => 4!/2! = 12 1,4,0,0 => 4!/2! = 12 0,1,2,2 => 4!/2! = 12 0,0,2,3 => 4!/2! = 12 Total = 4+4+ 12+12+12+12 = 56. 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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Re: Integers less than 10,000
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13 May 2010, 18:14
gurpreetsingh wrote: Pairs possible : 1,2,1,1 ; 1,3,1,0 ; 1,4,0,0 ; 0,1,2,2 ; 0,0,0,5 ; 0,0,2,3
which gives : 1,2,1,1 => 4!/3! = 4 0,0,0,5 =>=> 4!/3! = 4
1,3,1,0 => 4!/2! = 12 1,4,0,0 => 4!/2! = 12 0,1,2,2 => 4!/2! = 12 0,0,2,3 => 4!/2! = 12
Total = 4+4+ 12+12+12+12 = 56.
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: Quote: XXXXX = 2111 XXXXX = 0032 XXXXX = 0122 = 0122 XXXXX = 1004 = 1004 I hope it helps



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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
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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
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12 Oct 2010, 08:46
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.
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Re: How many positive integers less than 10,000 are there in
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16 Jun 2015, 23:59
Bunuel wrote: 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 from 0 to 9, 5 ( 1 number) from 10 to 99, we have 14,23,32,41,50 (5 numbers) from 100 to 999, we have 104,113. 122,131, 140, 203,212,221, 230,302, 311,320,401,410,500 (15 numbers) from 1000 to 9999, we have 1004,1040,1103,1112, 1121,1130,1202,1211,1220,1301,1310,1400, 2003,2012,2021,2030,2102,2111,2120,2201,2210,2300,3002.3011.3020,3101,3110,3200,4001,4010,4100,5000 (32 numbers) So we have only 53 numbers. Can anyone tell the numbers which I miss?



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Re: How many positive integers less than 10,000 are there in
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16 Apr 2018, 08:32
I don't like the sticks method. It is not intuitive at all. And I will no way even think of that in the exam. For me, the usual way is better. Sum of digits > 5 and Less than 10,000. (0,0,0,5)  4!/3!  4 (0,0,1,4)  4!/2!  12 (0,0,2,3)  4!/2!  12 (0,1,1,3)  4!/2!  12 (0,1,2,2)  4!/2!  12 (1,1,1,2)  4!/3!  4 Add them all > 56 Isnt this simple enough? And can be extrapolated easily to any question of this sort no? Bunuel wrote: 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



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Re: How many positive integers less than 10,000 are there in
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21 May 2019, 07:13
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 can we use the same method for any other number? Let's say 6?



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Re: How many positive integers less than 10,000 are there in
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21 May 2019, 07:15
AmanMatta wrote: 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 can we use the same method for any other number? Let's say 6?Please read the whole thread: https://gmatclub.com/forum/howmanypos ... ml#p798579
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Re: How many positive integers less than 10,000 are there in
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21 May 2019, 08:56
This is great concept and I had faced such question many times and failed again and again. SO let me clear this concept here. Suppose, I have 9 identical bananas which are to be distributed among 4 persons in such a way that a person can get no banana to 9 bananas. So here I have to allocate 9 items at 4 locations. Means I have 9 B and 3 separators. From your concept, BBBBBBBBBIII is to be arranged to get ways of distribution. 12!/ (9!3!) will give me the correct result. Please appraise my understanding so that I could not repeat the mistake for such problems. 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.



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Re: How many positive integers less than 10,000 are there in
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22 Sep 2019, 13:13
Bunuel wrote: 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 positive integers < 10,000… are from 1 to 9999, so we can go from 1digit number to 4digits number; \(x_1+x_2+x_3+x_4=5…(digit.slots)=(sum.five)\) \(c(n+r1,r1)=c(5+41,41)=c(8,3)=\frac{8!}{3!5!}=\frac{8,7,6}{3,2}=8•7=56\) Answer (C)




Re: How many positive integers less than 10,000 are there in
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