The posted question is virtually the same as the following:
How many ways can 6 identical chocolates be distributed among 4 children?

We can use the SEPARATOR METHOD.

6 identical chocolates are to be separated into -- at most -- 4 groupings.
Thus, we need 6 chocolates and 3 separators:
O|O|OO|OO

Every arrangement of the elements above represents one way to distribute the 6 chocolates among 4 children A, B, C and D:
O|O|OO|OO = A gets 1 chocolate, B gets 1 chocolate, C gets 2 chocolates. D gets 2 chocolates
This case is the equivalent of the following integer: 1122
||OO|OOOO = A gets 0 chocolates, B gets 0 chocolates, C gets 2 chocolates. D gets 4 chocolates
This case is the equivalent of the following integer: 0024 = 24
OOOOOOOOOO||| = A gets all 6 chocolates
This case is the equivalent of the following integer: 6000

To count the number of possible distributions, we need to calculate the number of ways to position the 3 separators.
Since there are 6 chocolates and 3 separators -- yielding a total of 9 positions -- the 3 separators must occupy 3 of the 9 positions.
From 9 positions, the number of ways to choose 3 to assign to the 3 separators = 9C3 \(= \frac{9*8*7}{3*2*1} = 84 \)



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I will break 6 into 4 digits first then find the number of satisfactory numbers for each set of numbers.
To break 6 down, we can have:
(1) 6 + 0 + 0 + 0. Note 0006 also counts, so this is 4 numbers. We can use 4! / (3!*1!) = 4 for all of these. (3! is to get rid of the duplicate 0's.)
(2) 5 + 1 + 0 + 0. 4!/(2!) = 12.
(3) 4 + 2 + 0 + 0. Similar as above, 12 numbers.
(4) 4 + 1 + 1 + 0. 4! / 2! = 12 numbers.
(5) 3 + 3 + 0 + 0. 4! / (2! * 2!) = 6.
(6) 3 + 2 + 1 + 0. 4! = 24.
(7) 3 + 1 + 1 + 1. 4! / 3! = 4.
(8) 2 + 2 + 2 + 0. 4! / 3! = 4.
(9) 2 + 2 + 1 + 1. 4! / (2!*2!) = 6.

In total, we have 4 + 36 + 6 + 24 + 8 + 6 = 40 + 30 + 14 = 84 numbers.

Another quick way to do this is to imagine a number that starts with 0000 and we are adding 1's in front of the slots to make the digits sum to 6, and we add 1's 6 times. For example 110, 110, 110, 0. Would correspond to 2220. 0, 0, 1110, 1110 would correspond to 0033. Regardless we can add 1's everywhere, so the first 9 digits (not including the very last 0 since we only add 1's in front of 0's) can be randomized. There are 6 one's and 3 zero's however so in total there are 9! / (6!*3!) = 9C3 = 9 * 8 * 7 / 6 = 3 * 4 * 7 = 84 combinations that are unique.

Ans: D
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This is same as donut problem.
Imagine that you have 6 identical donuts and want distribute among 4 mens, such that each men may get 0 to all 6 donuts.
It is permutation of 1,1,1,1,1,1,X,X,X => 9!/ (6! * 3!) = (9 * 8 * 7)/3! = 84

The question never really mentioned we need to restrict the solution for 4 digit integers only..Bunuel Am I Missing something here ?

You are right. We should count not only four-digit numbers, which have the sum of their digits equal to 6, but also three-, two-, and single-digit numbers. The solutions above do not miss this point and also account for these cases.
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