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Re: Integers less than 10,000 [#permalink]
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26 May 2013, 04:47
cumulonimbus wrote: 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. Yes, that's correct.
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Re: How many positive integers less than 10,000 are there in [#permalink]
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05 Jul 2013, 02:27



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Re: How many positive integers less than 10,000 are there in [#permalink]
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25 Nov 2013, 05:18
Excellent solution Bunuel. Saves a minute at minimum !!!



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Re: How many positive integers less than 10,000 are there in [#permalink]
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29 Nov 2013, 11:03
Hi Bunnel, please can explain when the separator concept is to be used and how to use it. Basically i did not understand in this question that why have we considered only 4 digit number. Please help.



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Re: How many positive integers less than 10,000 are there in [#permalink]
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29 Nov 2013, 11:09



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Re: How many positive integers less than 10,000 are there in [#permalink]
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27 Dec 2013, 15:33
Exceptionnal technique! Thanks all for this! Saves a lot of time in a lot of situations! Incredible minds! thanks !!
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Re: How many positive integers less than 10,000 are there in [#permalink]
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05 May 2014, 17:39
Bunuel's method was clearly simpler and faster, though I would hardly come up with a similar solution in the gmat.
I did it in a different way, can someone check if the approach was valid?
5 and 0s: 4P1*3C3 = 4*1 = 4 4, 1 and 0s: 4P1*3P1*2C2 = 4*3*1 = 12 3, 2 and 0s: 4P1*3P1*2C2 = 4*3*1 = 12 2, 2, 1 and 0: 4P2*2P1*1 = 12*2 = 24 2, 1, 1, 1: 4P1*3C3 = 4*1 = 4
4 + 12+ 12 + 24 + 4 = 56



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Re: How many positive integers less than 10,000 are there in [#permalink]
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30 Nov 2014, 14:45
Hey Bunel would you plz tell me in this formula 8! / (5!*3!) ,you got the numbers 8,5, and 3 from where ???



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How many positive integers less than 10,000 are there in [#permalink]
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15 Dec 2014, 05:33
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 Bunuel, Could you please clarify why we are taking 5 d's and 3 seprator (/). i am getting confusion here. we can take four separator also and get the result. Thanks.



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Re: How many positive integers less than 10,000 are there in [#permalink]
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26 Dec 2014, 04:43
Hi,
If I may venture to propose the solution I used and you can tell me what I am missing.
I started by testing it from 010. There is one such number (5). From 1120 there is one such number (14). This led me realize than from 099 there are 9 such numbers. So, this was the lengthy part of my thinking process (not lengthy at all).
From 099: 9 numbers. From : 100999: 9*2= 18 numbers From 10009999: 9*3= 27 numbers Adding them up: 9+18+27= 54.
This is close enough so I decided to choose 56 anyway, but since there are 2 numbers missing, could you tell me why and where?
Thank you, Natalia



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Re: How many positive integers less than 10,000 are there in [#permalink]
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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 [#permalink]
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19 May 2016, 00:21
As it says integers less than 100,000,why only four digit numbers are considered ? Why not three digits,two digits and single digit integers considered?



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Re: How many positive integers less than 10,000 are there in [#permalink]
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19 May 2016, 03:34



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Re: How many positive integers less than 10,000 are there in [#permalink]
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16 Apr 2017, 14:47
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 hello, i complete understand the formula, but what i still do not understand is how why there are only 5 numbers? It should be 6 if o is included



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Re: How many positive integers less than 10,000 are there in [#permalink]
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21 Jul 2017, 10:10
X1 + X2 +X3 +X4 =5
The number of solutions of this equation for X1 X2, X3 and X4>=0 : n+r1(C)r1
Here r=4 and n=5 Hence solution: 8C3



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Re: How many positive integers less than 10,000 are there in [#permalink]
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06 Sep 2017, 22:41
walker wrote: 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 ToolKitHi walker, Can you please explain this?



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Re: How many positive integers less than 10,000 are there in [#permalink]
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04 Oct 2017, 05:30
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: \({n+r1}_C_{r1}={5+41}_C_{41}={8}C3=\frac{8!}{5!3!}=56\) Hi Bunuel, I just want to make sure that i understood the concept. Let us assume that the question stem ask for a sum of 4 instead of 5. Will the answer be: XXXXIII i.e. 7!/(3!x4!)? If the question asks for a sum of five for numbers below 20,000 will the answer be 9!/(5!x4!)? if the question asks for a sum of four for numbers below 20,000 will the answer be : XXXX0IIII i.e. 9!/(4!x4!)? Another thing, is there a formula if you want to distribute n different objects among k people? (i could count the case when n and k are small for example n=3 and k=2 but i was wondering if there was a general formula for that)




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