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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.

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.

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: \(|dd|d|dd\) 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

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.

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 0-10. There is one such number (5). From 11-20 there is one such number (14). This led me realize than from 0-99 there are 9 such numbers. So, this was the lengthy part of my thinking process (not lengthy at all).

From 0-99: 9 numbers. From : 100-999: 9*2= 18 numbers From 1000-9999: 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?

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

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18 Jan 2015, 07:52

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:

XX|X|X|X = 2111 ||XXX|XX = 0032

etc.

There are 8C3 ways to determine where to place the separators. Hence, the answer is 8C3 = 56.

Sorry i could not understand how you have started with 3 Separators , why not 4 or 5 ?
_________________

Thanks, Lucky

_______________________________________________________ Kindly press the to appreciate my post !!

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?

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?

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?

They are considered. If you read the whole thread you'll find the answer to your question.
_________________

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: \(|dd|d|dd\) 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

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

gmatclubot

Re: How many positive integers less than 10,000 are there in
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16 Apr 2017, 14:47

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