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

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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
shaktirdas19 wrote:
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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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 $$\frac{8!}{5!3!}=56$$.

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: $${n+r-1}_C_{r-1}={5+4-1}_C_{4-1}={8}C3=\frac{8!}{5!3!}=56$$

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

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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+r-1(C)r-1

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 whi [#permalink]

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09 Aug 2017, 22:05
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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 0000-9999 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.

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Hi walker,

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Re: How many positive integers less than 10,000 are there in   [#permalink] 06 Sep 2017, 22:41

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