Different ways to solve this please!

[winniethepooh]

2. 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

I want to learn the Partition/seperation method if any one can comment on that!(seems to be a good trick!)

[GMATGuruNY]

2. 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

I want to learn the Partition/seperation method if any one can comment on that!(seems to be a good trick!)

Check here:

https://www.beatthegmat.com/experts-any- ... 82307.html

[winniethepooh]

Thanks for prompt reply Mitch!
I love that method.
Mitch, could you please correct me if I am wrong:
In any question which can be solved with separators, the number of separators = the number of minimum elements - 1?(like in the 4 balls among 3 children problem)

[knight247]

Hey Mitch,
Great explanation. Just need one more explanation though. What if the question were modified to

How many positive integers less than 10,000 are there in which sum of digits equals 3?

Could you help with the explanation for that? Appreciate it

[goalevan]

How many positive integers less than 10,000 are there in which sum of digits equals 5?

We know that a number less than 10,000 contains up to 4 digits that must sum to 5.

The combinations that sum to 5:

5,0
4,1
3,2
3,1,1
2,2,1
2,1,1,1

This creates anagrams for the 4-digits of the integers as follows:

5000, 0500, 0050, 0005 can be represented by AXXX = 4C1 = 4C4 = 4!/3! = 4
4100, 1400, 4010, etc.. ABXX = 4!/2! = 12
3200, 2300, etc.. ABXX = 4!/2! = 12
3110, 1310, etc.. ABXX = 4!/2! = 12
2210, 2120, etc.. ABXX = 4!/2! = 12
2111, 1211, etc.. AXXX = 4!/3! = 4

12*4 + 4*2 = 48 + 8 = 56 integers whose digits sum to 5.

How many positive integers less than 10,000 are there in which sum of digits equals 3?

The combinations that sum to 3:

3,0
2,1
1,1,1

This creates anagrams for the 4-digits of the integers as follows:

3000, 0300, etc.. AXXX = 4!/3! = 4
2100, 2010, etc.. ABXX = 4!/2! = 12
1110, 1011, etc.. AXXX = 4!/3! = 4

12*1 + 4*2 = 12 + 8 = 20 integers whose digits sum to 3.

Using Mitch's separator method:

DDD||| for 3,0,0,0
DD|D|| for 2,1,0,0
D|D|D| for 1,1,1,0

We compute the anagram for |||DDD

6C3 = 6!/(3!3!) = 6*5*4 / 3*2 = 5*4 = 20

[GMATGuruNY]

Hey Mitch,
Great explanation. Just need one more explanation though. What if the question were modified to

How many positive integers less than 10,000 are there in which sum of digits equals 3?

Could you help with the explanation for that? Appreciate it

Using the separator method:

Let UUU represent the 3 units.
Since the 3 units are to be distributed among up to 4 digits, we need 3 separators.
Let ||| represent the 3 separators.
Number of ways to arrange the 6 elements UUU||| = 6!/3!3! = 20.

Since there are not many ways in which 1 to 4 digits could have a sum of 3, we simply could count the different possibilities:

Number of ways to arrange the digits 0003 = 4!/3! = 4.
Number of ways to arrange the digits 0012 = 4!/2! = 12.
Number of ways to arrange the digits 0111 = 4!/3! = 4.
Total number of ways = 4+12+4 = 20.

[sss2534]

Hey Mitch,
Great explanation. Just need one more explanation though. What if the question were modified to

How many positive integers less than 10,000 are there in which sum of digits equals 3?

Could you help with the explanation for that? Appreciate it

Using the separator method:

Let UUU represent the 3 units.
Since the 3 units are to be distributed among up to 5 digits, we need 4 separators.
Let |||| represent the 4 separators.
Number of ways to arrange the 7 elements UUU|||| = 7!/3!4! = 35.

Since there are not many ways in which 1 to 5 digits could have a sum of 3, we simply could count the different possibilities:

Number of ways to arrange the digits 00003 = 5!/4! = 5.
Number of ways to arrange the digits 00012 = 5!/3! = 20.
Number of ways to arrange the digits 00111 = 5!/2!3! = 10.
Total number of ways = 5+20+10 = 35.

Hey Mitch..

The question says "how many integers less than 10,000" -- so it should be 3 separators and 4 digits, correct?

Using the stars and bars counting method:

There are 4 elements (0,1,2,3) --> that need to be distributed into 4 bins (3 separators).

n = 4 (total number of elements)

k = 3 (total number of separators)

Total number of distributions = (n+ k -1) C (k)

6C3 = 20;

Your solution assumes that there are 4 separators (but the question only asks for digits less than 10,000). Are you sure 35 is the correct solution?

[GMATGuruNY]

Your solution assumes that there are 4 separators (but the question only asks for digits less than 10,000). Are you sure 35 is the correct solution?

You are correct. I've amended my post above. I had counted the number of integers less than 100,000 the sum of whose digits is 3. Thanks for the heads up.