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"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:

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 \({n+r-1}_C_{r-1}\).

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.

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.

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 \({n+r-1}_C_{r-1}\).

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.

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.

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

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12 Jul 2013, 22:44

alphabeta1234 wrote:

Bunuel,

Many apologies for reviving this method. But I gotta say I love your "stars and bars" method. My Question is it seems to only work for this question if the sum is less than 9 for this question. You can't have more than 9 stars in each slot. What if the question had asked,

How many positive integers less than 10,000 are there in which the sum of the digits equals 13?

Now this is a little tricker. Here is another example from another problem posted on this forum that illustrates the problem. Is there a way we can exand "stars and bars" to numbers that sum larger than 9?

A wheel of fortune contains numerical values from 1 to 8. The scoring system of the game is based on the sum of these values. If the host were to spin the wheel three times, how many possible number of combinations are there that will give the player the sum of 16 points?

Bunuel,

Like alphabeta1234 mentioned above, there is a limitation to the stars and bars method. How then would we be able to find out the number of positive integers less than 10,000 if their sum is to be, say, 13? He mentioned the wheel of fortune method, but looking it up on the internet didn't turn up much that was useful.

Many apologies for reviving this method. But I gotta say I love your "stars and bars" method. My Question is it seems to only work for this question if the sum is less than 9 for this question. You can't have more than 9 stars in each slot. What if the question had asked,

How many positive integers less than 10,000 are there in which the sum of the digits equals 13?

Now this is a little tricker. Here is another example from another problem posted on this forum that illustrates the problem. Is there a way we can exand "stars and bars" to numbers that sum larger than 9?

A wheel of fortune contains numerical values from 1 to 8. The scoring system of the game is based on the sum of these values. If the host were to spin the wheel three times, how many possible number of combinations are there that will give the player the sum of 16 points?

Bunuel,

Like alphabeta1234 mentioned above, there is a limitation to the stars and bars method. How then would we be able to find out the number of positive integers less than 10,000 if their sum is to be, say, 13? He mentioned the wheel of fortune method, but looking it up on the internet didn't turn up much that was useful.

There is a direct formula given on previous page.
_________________

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

Show Tags

13 Jul 2013, 00:52

Bunuel wrote:

keylimepie wrote:

alphabeta1234 wrote:

Bunuel,

Many apologies for reviving this method. But I gotta say I love your "stars and bars" method. My Question is it seems to only work for this question if the sum is less than 9 for this question. You can't have more than 9 stars in each slot. What if the question had asked,

How many positive integers less than 10,000 are there in which the sum of the digits equals 13?

Now this is a little tricker. Here is another example from another problem posted on this forum that illustrates the problem. Is there a way we can exand "stars and bars" to numbers that sum larger than 9?

A wheel of fortune contains numerical values from 1 to 8. The scoring system of the game is based on the sum of these values. If the host were to spin the wheel three times, how many possible number of combinations are there that will give the player the sum of 16 points?

Bunuel,

Like alphabeta1234 mentioned above, there is a limitation to the stars and bars method. How then would we be able to find out the number of positive integers less than 10,000 if their sum is to be, say, 13? He mentioned the wheel of fortune method, but looking it up on the internet didn't turn up much that was useful.

There is a direct formula given on previous page.

I'm sorry. I'm afraid I don't see which one I can apply to the situation. Could you please point me in the right direction?

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]

Show Tags

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

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