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How many positive integers less than 10,000 are there in [#permalink]
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13 Oct 2009, 19:37
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Re: Integers less than 10,000 [#permalink]
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14 Oct 2009, 11:23
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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 [#permalink]
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14 Oct 2009, 19:06
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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 [#permalink]
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11 Jan 2010, 12:15
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 [#permalink]
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11 Jan 2010, 15:46
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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 [#permalink]
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11 Jan 2010, 17:17
What's the concept behind the separators?



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Re: Tough Combinatronics questions [#permalink]
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11 Jan 2010, 22:58
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Re: Tough Combinatronics questions [#permalink]
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11 Jan 2010, 23:08
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Bunuel wrote: 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: \({n+r1}_C_{r1}={5+41}_C_{41}={8}C3=\frac{8!}{5!3!}=56\) That was brilliant. Wow.
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Re: Integers [#permalink]
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10 Mar 2010, 03:23
Grt explanation Bunuel.



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Re: Integers [#permalink]
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10 Mar 2010, 22:31
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 [#permalink]
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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 65549 times ]
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Re: Integers less than 10,000 [#permalink]
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07 Apr 2010, 05:40
Fantastic explanation. +1



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Re: Integers less than 10,000 [#permalink]
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amazing minds I solved it using a lengthy process and still got it wrong. Don't know what I did wrong ...well does not matter.



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Re: Integers less than 10,000 [#permalink]
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10 Apr 2010, 02:52
Thanks Bunuel and AKProdigy +1 to you both excellent approach.
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Re: Integers less than 10,000 [#permalink]
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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 [#permalink]
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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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Re: Integers less than 10,000 [#permalink]
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14 May 2010, 10:43
great explaination & good question thanks Bunuel



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Re: Integers less than 10,000 [#permalink]
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17 May 2010, 10:34
Bunuel, Great explanation! Can you provide another example of a digit problem where we can apply this concept?



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Re: Integers less than 10,000 [#permalink]
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18 May 2010, 18:21
Quote: 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



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Re: Integers less than 10,000 [#permalink]
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14 Jun 2010, 08:37
bibha wrote: HEy, How do we know how many separators to use? I mean how did we get 8 and 3?? Hi Bibha! Have you read the previous explanations? 8 because we have 8 elements 5 + 3 and 3 because we need a four digit number I hope it helps




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