I agree with Brent that the posted problem is beyond the scope of the GMAT.
That said, here is a similar problem, along with my solution:
How many of the four-digit numbers with non-zero digits have the sum of their digits as 12?
165
330
132
440
170
A sum of 12 units is to be DISTRIBUTED among 4 digits.
None of the digits can be greater than 9.
This is the equivalent of the following problem:
How many ways can 12 identical chocolates be distributed among 4 children A, B, C and D if no child can receive more than 9 chocolates?
To guarantee that no child receives more than 9 chocolates, first distribute 1 chocolate to each child.
Now count the number of ways to distribute the REMAINING 8 CHOCOLATES to the 4 children.
The following is called the SEPARATOR method.
8 identical chocolates are to be separated into -- at most -- 4 groupings.
Thus, we need 8 chocolates and 3 separators:
OO|OO|OO|OO
Each arrangement of the elements above represents one way to distribute the 8 chocolates among 4 children A, B, C and D:
OO|OO|OO|OO = A gets 2 chocolates, B gets 2 chocolates, C gets 2 chocolates, D gets 2 chocolates.
OOOOOO|||OO = A gets 6 chocolates, D gets 2 chocolates.
OOOOOOOO||| = A gets all 8 chocolates.
And so on.
To count all of the possible distributions, we simply need to count the number of ways to arrange the 11 elements above (the 8 identical chocolates and the 3 identical separators).
The number of ways to arrange 11 elements = 11!.
But when an arrangement includes identical elements, we must divide by the number of ways to arrange the identical elements.
The reason:
When the identical elements swap positions, the arrangement doesn't change, reducing the total number of unique arrangements.
Thus, we must divide by 8! (the number of ways to arrange the 8 identical chocolates) and by 3! (the number of ways to arrange the 3 identical separators):
11!/(8!3!) = 165.
The correct answer is
A.
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