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Bunuel
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How many positive integers less than 10,000 are there in Updated on: 13 Oct 2009, 21:27
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.
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

95% (hard)

Question Stats:

43% (02:43) correct 57% (03:01) wrong based on 1976 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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C
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AKProdigy87
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How many positive integers less than 10,000 are there in 14 Oct 2009, 12:23
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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:

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.
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How many positive integers less than 10,000 are there in 07 Apr 2010, 03:50
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Ramsay
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
Show more

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
pTNfS-2e270de4ca223ec2741fa10b386c7bfe.jpg [ 63.83 KiB | Viewed 130754 times ]
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How many positive integers less than 10,000 are there in 13 May 2010, 16:08
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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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How many positive integers less than 10,000 are there in 14 Oct 2009, 20:06
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AKProdigy87
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.
Show more

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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How many positive integers less than 10,000 are there in 11 Jan 2010, 16:46
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zaarathelab
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
Show more

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: \({n+r-1}_C_{r-1}={5+4-1}_C_{4-1}={8}C3=\frac{8!}{5!3!}=56\)
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How many positive integers less than 10,000 are there in 11 Jan 2010, 18:17
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What's the concept behind the separators?
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How many positive integers less than 10,000 are there in 11 Jan 2010, 23:58
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mrblack
What's the concept behind the separators?
Show more

This is what I found in the net about this question explaining the concept:
Attachment:
pTNfS-2e270de4ca223ec2741fa10b386c7bfe.jpg
pTNfS-2e270de4ca223ec2741fa10b386c7bfe.jpg [ 63.83 KiB | Viewed 72279 times ]

Hope it's clear.
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How many positive integers less than 10,000 are there in 13 May 2010, 18:14
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gurpreetsingh
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.
Show more
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
Show more

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

I hope it helps
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How many positive integers less than 10,000 are there in 12 Oct 2010, 08:35
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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?
thanks in advance
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anilnandyala
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
Show more

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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How many positive integers less than 10,000 are there in 16 Apr 2018, 08:32
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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

Add them all --> 56

Isnt this simple enough? And can be extrapolated easily to any question of this sort no?

Bunuel
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
Show more
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How many positive integers less than 10,000 are there in 13 Jan 2020, 14:41
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Bunuel
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
Show more


If the sum of digits of a number is equal to 5, then each digit must be less than or equal to 5. Furthermore, we can consider each number as a “4-digit” number. For example, 5 can be considered as 0005 and 104 can be considered as 0104. We can classify the numbers into cases according to the largest digit of the number.

1) If the largest digit is 5, then the other three digits must be all 0. The number of ways to arrange one 5 and three 0’s is 4!/3! = 4.

2) If the largest digit is 4, then the other three digits must consist of one 1 and two 0’s. The number of ways to arrange one 4, one 1 and two 0’s is 4!/2! = 4 x 3 = 12.

3) If the largest digit is 3, then the other three digits must consist of one 2 and two 0’s OR two 1’s and one 0. The number of ways to arrange one 3, one 2 and two 0’s is 4!/2! = 12, and the number of ways to arrange one 3, two 1’s and one 0 is also 4!/2! = 12. So the total number of ways in this case is 12 + 12 = 24.

4) If the largest digit is 2, then the other three digits must consist of all 1 OR one 2, one 1 and one 0. The number of ways to arrange one 2 and three 1’s is 4!/3! = 4, and the number of ways to arrange two 2’s, one 1 and one 0 is also 4!/2! = 12. So the total number of ways in this case is 4 + 12 = 16.

Since the largest digit of the number can’t be 1 or 0, the total number of ways in which the sum of the digits is equal to 5 is 4 + 12 + 24 + 16 = 56.

Answer: C
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How many positive integers less than 10,000 are there in 12 Mar 2024, 22:12
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CAN WE SOLVE THIS QUESTION WITH THE METHOD WE HAVE USED ABOVE?­A certain company assigns employees to offices in such a way that some of the offices can be empty and more than one employee can be assigned to an office. In how many ways can the company assign 3 employees to 2 different offices?


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How many positive integers less than 10,000 are there in 13 Mar 2024, 00:13
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CAN WE SOLVE THIS QUESTION WITH THE METHOD WE HAVE USED ABOVE?­A certain company assigns employees to offices in such a way that some of the offices can be empty and more than one employee can be assigned to an office. In how many ways can the company assign 3 employees to 2 different offices?



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­
d's we distribute in the original question are identical, whereas the individuals mentioned in the question you referenced are not. Hence, we must account for permutations of different individuals there.

For a detailed discussion on this question, please visit:

https://gmatclub.com/forum/a-certain-co ... 88936.html
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How many positive integers less than 10,000 are there in 27 Apr 2025, 11:23
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Hi Bunuel.

I see that this topic is very old, but there is something that I don't fully understand in this question

I arrived to this question by accident (while checking this other question's replies: https://gmatclub.com/forum/what-is-the- ... 05993.html)

How come are there only 3 separators and not 4 (ddddd|||)? If there were 4 separators i could have a number such as: |dd|d|d|d (2 1 1 1) there would still be 4 digit (<10,000) and the sum would still be 5.

What am I missing here? in the explanation you posted in the picture with selecting 3 people out of 8, it is also tricky for me.

I would really appreciate your help.

Thanks in advance.

Regards,
Juan Diego

Bunuel
Ramsay
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
Show more
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How many positive integers less than 10,000 are there in 27 Apr 2025, 12:00
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jdtg2610
Hi Bunuel.

I see that this topic is very old, but there is something that I don't fully understand in this question

I arrived to this question by accident (while checking this other question's replies: https://gmatclub.com/forum/what-is-the- ... 05993.html)

How come are there only 3 separators and not 4 (ddddd|||)? If there were 4 separators i could have a number such as: |dd|d|d|d (2 1 1 1) there would still be 4 digit (<10,000) and the sum would still be 5.

What am I missing here? in the explanation you posted in the picture with selecting 3 people out of 8, it is also tricky for me.

I would really appreciate your help.

Thanks in advance.

Regards,
Juan Diego

Bunuel
Ramsay
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
Show more

You can have digits before the first separator. For example, 2111 is dd|d|d|d. We need only 3 separators to split into 4 digits.

Similar questions to practice:
https://gmatclub.com/forum/larry-michae ... 08739.html
https://gmatclub.com/forum/12-marbles-a ... 09082.html
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Hope this helps.
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Bunuel

Thanks for the reply. That helped.
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