Problem Solving

How many positive integers less than 10,000 are there in which sum of digits equals 5?
(a) 31
(b) 51
(c) 56
(d) 62
(e) 93

Chaitanya_1986 wrote:How many positive integers less than 10,000 are there in which sum of digits equals 5?
(a) 31
(b) 51
(c) 56
(d) 62
(e) 93

Use the separator method.
Let sum of 5 = DDDDD.

We can use these five D's to represent any integer in which the sum of the digits is 5:
DDDDD = 5.
DDDD D = 41.
D DDD D = 131.

To be less than 10,000, the integer can have at most 4 digits.
Thus, the five D's can be separated at most into 4 groups. For example:
D D D DD = 1112.

Thus, there are at most 3 separators dividing the five D's.
Let the 3 separators = |||.

Thus:
|||DDDDD = 5.
||DDDD|D = 41.
|D|DDD|D = 131.
D|D|D|DD = 1112.

Thus, to count the number of integers, we need to count the number of ways to arrange DDDDD|||.
Number of ways to arrange DDDDD||| = 8!/(5!3!) = 56.

The correct answer is C.

Hi Mitch,

Thanks a lot for your explanation.

Could you please tell me When can we use a separator method? and in which type of problems can we use this method.

Chaitanya_1986 wrote:Hi Mitch,

Thanks a lot for your explanation.

Could you please tell me When can we use a separator method? and in which type of problems can we use this method.

The separator method can used to count the number of ways to distribute identical elements.

Four identical balls are thrown into the air. Three children are playing a game in which each child holds a barrel that is to be used to catch the balls. If at the end of the game all of the balls have been caught, in how many ways could the four balls be distributed among the three children?

Using the separator method, we'd have four balls (BBBB) and two separators (||), since the balls can be split into a maximum of 3 divisions.

Number of ways to arrange BBBB|| = 6!/(4!2!) = 15.

Hi Mitch:

Further on the example that you've given below, let's say I want to do this using normal P n C.

Case 1) One child catches 2 balls and remaining 2 children catch 1 ball each.

Case 2) One child catches 3 balls and out of remaining 2 children, 1 catches 1 ball.

Case 3) One child catches all 4 balls.


So, for Case 1): Choosing 1 child out of 3 to catch 2 balls - 3C1. This child can catch the ball in 4!/2! ways
Choosing 2 children out of 2 to catch 1 ball each - 2C2*1

Total no. of ways = 3C1*4!/2!*2C1*1 = 36

Similarly we can calculate combinations for Case 2 and 3. Of course my answer doesn't match yours but can you please point out what's wrong with this approach?

prakhag wrote:Hi Mitch:

Further on the example that you've given below, let's say I want to do this using normal P n C.

Case 1) One child catches 2 balls and remaining 2 children catch 1 ball each.

Case 2) One child catches 3 balls and out of remaining 2 children, 1 catches 1 ball.

Case 3) One child catches all 4 balls.


So, for Case 1): Choosing 1 child out of 3 to catch 2 balls - 3C1. This child can catch the ball in 4!/2! ways
Choosing 2 children out of 2 to catch 1 ball each - 2C2*1

Total no. of ways = 3C1*4!/2!*2C1*1 = 36

Similarly we can calculate combinations for Case 2 and 3. Of course my answer doesn't match yours but can you please point out what's wrong with this approach?

If you revisit my post above, you'll note that I've edited the question to make its intention clearer.

The separator method counts the number of ways in which the four balls can be distributed among the 3 children:
Number of ways to arrange BBBB|| = 6!/4!2! = 15.

Here are all the possible distributions:
1 child catches all 4 balls, the other 2 children each catch 0 balls:
Number of ways to arrange 4-0-0 = 3.

1 child catches 3 balls, 1 child catches 1 ball, 1 child catches 0 balls:
Number of ways to arrange 3-1-0 = 6.

2 children each catch 2 balls, 1 child catches 0 balls:
Number of ways to arrange 2-2-0 = 3.

1 child catches 2 balls, the other 2 children each catch 1 ball:
Number of ways to arrange 2-1-1 = 3.

Total number of ways to distribute the balls = 3+6+3+3 = 15.

Hi Mitch, won't the sequence of numbers matter in this case? For e.g. (14,41), (23,32), (1112, 1211,1121..)?

Thanks

GMATGuruNY wrote:

Chaitanya_1986 wrote:How many positive integers less than 10,000 are there in which sum of digits equals 5?
(a) 31
(b) 51
(c) 56
(d) 62
(e) 93

Use the separator method.

Let the 5 digits whose sum is 5 = DDDDD.

Any division of these five D's will yield an expression that represents a 5-digit integer, the sum of whose digits is 5:
DDDDD = 5.
DDDD D = 41.
D DDD D = 131.

To be less than 10,000, the integer can have at most 4 digits.
Thus, the five D's can be separated at most into 4 groups. For example:
D D D DD = 1112.

Thus, there are at most 3 separators dividing the five D's.
Let the 3 separators = |||.

Thus:
|||DDDDD = 5.
||DDDD|D = 41.
|D|DDD|D = 131.
D|D|D|DD = 1112.

Thus, to count the number of integers, we need to count the number of ways to arrange DDDDD|||.
Number of ways to arrange DDDDD||| = 8!/(5!3!) = 56.

The correct answer is C.

apex231 wrote:Hi Mitch, won't the sequence of numbers matter in this case? For e.g. (14,41), (23,32), (1112, 1211,1121..)?

Thanks

GMATGuruNY wrote:

Chaitanya_1986 wrote:How many positive integers less than 10,000 are there in which sum of digits equals 5?
(a) 31
(b) 51
(c) 56
(d) 62
(e) 93

Use the separator method.

Let the 5 digits whose sum is 5 = DDDDD.

Any division of these five D's will yield an expression that represents a 5-digit integer, the sum of whose digits is 5:
DDDDD = 5.
DDDD D = 41.
D DDD D = 131.

To be less than 10,000, the integer can have at most 4 digits.
Thus, the five D's can be separated at most into 4 groups. For example:
D D D DD = 1112.

Thus, there are at most 3 separators dividing the five D's.
Let the 3 separators = |||.

Thus:
|||DDDDD = 5.
||DDDD|D = 41.
|D|DDD|D = 131.
D|D|D|DD = 1112.

Thus, to count the number of integers, we need to count the number of ways to arrange DDDDD|||.
Number of ways to arrange DDDDD||| = 8!/(5!3!) = 56.

The correct answer is C.

The separator method is used to count the number of ways in which identical elements can be distributed.
Since in the problem above the sum of the digits must be 5, the separator method can be used to count the number of ways in which the 5-unit sum (DDDDD) can be distributed among the 1,2,3 or 4 digits.

To illustrate:
|D|DD|DD = a distribution of 0-1-2-2, representing the number 122.
|DD|D|DD = a distribution of 0-2-1-2, representing the number 212.
|DD|DD|D = a distribution of 0-2-2-1, representing the number 221.

Notice that the distributions above account for all the different ways in which the digits 1,2,2 can be ordered in a 3-digit integer.

Thus, by using the separator method to count all the different ways in which the 5-unit sum can be distributed, we account for every possible sequence of digits.

Hii Mitch....

Thanks for the explanation ...can u plz suggest me books for quantative from where i can clear basics and move to advance problems(with high difficulty) as i m looking to score above 700....kindly suggest...

Regards,
Macky

That was great Mitch! I understand how to use the separator method for distribution problems now, but I would love to know more on the underlying math and reasons behind it. Do you have any links that explores this subject in more detail? I tried to search for "separator method math" and other combinations in Google, but can't find anything.

Thanks again!

I think I got it now. The separator method is an ingenious way to calculate this problem. For more information check out:
https://gmatclub.com/forum/integers-less-than-85291.html

(Be sure to scroll down to the example with the stick figures.)

just one thing to clear...how do we get the number of separators????

nafiul9090 wrote:just one thing to clear...how do we get the number of separators????

Number of separators = Maximum number of distributions - 1.

In the digits problem above, the sum can be distributed among up to 4 digits, so the maximum number of distributions is 4.
Number of separators = 4-1 = 3.

In the game problem that I posted, the balls can be distributed among up to 3 children, so the maximum number of distributions is 3.
Number of separators = 3-1 = 2.

Other variants of this : (Not an official question-Just a thought of mind)

In how many ways can 10 chocolates be divided among 6 children so all chocolates may go to a single child ?
CCCCCCCCCC||||| = 15! / 10!5! (I hope this is correct ? )


How to solve this ?

In how many ways can 10 chocolates be divided among 6 children so that each child gets at least 1 chocolate ?

GMATGuruNY wrote:

apex231 wrote:Hi Mitch, won't the sequence of numbers matter in this case? For e.g. (14,41), (23,32), (1112, 1211,1121..)?

Thanks

GMATGuruNY wrote:

Chaitanya_1986 wrote:How many positive integers less than 10,000 are there in which sum of digits equals 5?
(a) 31
(b) 51
(c) 56
(d) 62
(e) 93

Use the separator method.

Let the 5 digits whose sum is 5 = DDDDD.

Any division of these five D's will yield an expression that represents a 5-digit integer, the sum of whose digits is 5:
DDDDD = 5.
DDDD D = 41.
D DDD D = 131.

To be less than 10,000, the integer can have at most 4 digits.
Thus, the five D's can be separated at most into 4 groups. For example:
D D D DD = 1112.

Thus, there are at most 3 separators dividing the five D's.
Let the 3 separators = |||.

Thus:
|||DDDDD = 5.
||DDDD|D = 41.
|D|DDD|D = 131.
D|D|D|DD = 1112.

Thus, to count the number of integers, we need to count the number of ways to arrange DDDDD|||.
Number of ways to arrange DDDDD||| = 8!/(5!3!) = 56.

The correct answer is C.

The separator method is used to count the number of ways in which identical elements can be distributed.
Since in the problem above the sum of the digits must be 5, the separator method can be used to count the number of ways in which the 5-unit sum (DDDDD) can be distributed among the 1,2,3 or 4 digits.

To illustrate:
|D|DD|DD = a distribution of 0-1-2-2, representing the number 122.
|DD|D|DD = a distribution of 0-2-1-2, representing the number 212.
|DD|DD|D = a distribution of 0-2-2-1, representing the number 221.

Notice that the distributions above account for all the different ways in which the digits 1,2,2 can be ordered in a 3-digit integer.

Thus, by using the separator method to count all the different ways in which the 5-unit sum can be distributed, we account for every possible sequence of digits.

Hi!

I understood the separators part, but can you please explain again, why are we taking 5Digits? What about integers of less then 100000?
Thank you!