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Math Expert V
Joined: 02 Sep 2009
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How many positive integers less than 10,000 are there in  [#permalink]

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Question Stats: 40% (02:44) correct 60% (02:57) wrong based on 913 sessions

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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.
Math Expert V
Joined: 02 Sep 2009
Posts: 59033
Re: Integers less than 10,000  [#permalink]

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28
30
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 [ 63.83 KiB | Viewed 81826 times ]

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Re: Integers less than 10,000  [#permalink]

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83
1
36
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.
##### General Discussion
Math Expert V
Joined: 02 Sep 2009
Posts: 59033
Re: Integers less than 10,000  [#permalink]

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6
9
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.

This is correct. Also this is the best way to solve this question. +1. (Solved exactly the same way)

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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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Joined: 02 Sep 2009
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6
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: $$|dd|d|dd$$ 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: $${n+r-1}_C_{r-1}={5+4-1}_C_{4-1}={8}C3=\frac{8!}{5!3!}=56$$
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What's the concept behind the separators?
Math Expert V
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mrblack wrote:
What's the concept behind the separators?

Attachment: pTNfS-2e270de4ca223ec2741fa10b386c7bfe.jpg [ 63.83 KiB | Viewed 27458 times ]

Hope it's clear.
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GMAT 1: 750 Q50 V40 ### Show Tags

2
1
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. Posted from GMAT ToolKit
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GMAT 1: 670 Q49 V31 GMAT 2: 710 Q50 V35 Re: Integers less than 10,000  [#permalink]

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22
13
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:
XX|X|X|X = 2111
||XXX|XX = 0032

|X|XX|XX = 0|1|2|2 = 0122
X|||XXXX = 1|0|0|4 = 1004

I hope it helps
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zestzorb wrote:
The method is called "stars and bars"

Thanks. I looked this up and am slightly confused when to use the (n-1)/(k-1) vs (n+k-1)/k or (n+k-1)/(n-1)

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  [#permalink]

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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?
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Joined: 02 Sep 2009
Posts: 59033
Re: Integers less than 10,000  [#permalink]

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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?

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 $${n+r-1}_C_{r-1}$$.

In our case we'll get: $${n+r-1}_C_{r-1}={6+4-1}_C_{4-1}={9}C3=\frac{9!}{6!3!}$$.

Hope it's clear.
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Re: How many positive integers less than 10,000 are there in  [#permalink]

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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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1
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

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  [#permalink]

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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

can we use the same method for any other number? Let's say 6?
Math Expert V
Joined: 02 Sep 2009
Posts: 59033
Re: How many positive integers less than 10,000 are there in  [#permalink]

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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: $$|dd|d|dd$$ 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

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  [#permalink]

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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?

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 $${n+r-1}_C_{r-1}$$.

In our case we'll get: $${n+r-1}_C_{r-1}={6+4-1}_C_{4-1}={9}C3=\frac{9!}{6!3!}$$.

Hope it's clear.
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Re: How many positive integers less than 10,000 are there in  [#permalink]

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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 1-digit number to 4-digits number;
$$x_1+x_2+x_3+x_4=5…(digit.slots)=(sum.five)$$
$$c(n+r-1,r-1)=c(5+4-1,4-1)=c(8,3)=\frac{8!}{3!5!}=\frac{8,7,6}{3,2}=8•7=56$$ Re: How many positive integers less than 10,000 are there in   [#permalink] 22 Sep 2019, 13:13
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