parveen110 wrote:Find the number of integral solutions to |x|+|y|+|z|=15.
a. 902
b. 728
c. 734
d. 904
e. 815
OA: A
I agree with Brent: this problem is WAY beyond the scope of the GMAT.
That being said, here's a solution that employs the SEPARATOR method.
For an explanation of this method, check my posts here:
https://www.beatthegmat.com/lets-have-a- ... 69973.html
https://www.beatthegmat.com/inserting-st ... 67423.html
https://www.beatthegmat.com/p-c-no-of-in ... 73465.html
https://www.beatthegmat.com/combinatorics-t124504.html
Note the following:
Because the problem above involves |x|, |y| and |z|, any positive option for x, y, or z implies a corresponding negative option.
For example, if x=10 is an option, so is x=-10.
Case 1: All 3 integers are nonzero
Here, a sum of 15 is to be distributed among 3 nonzero integers.
Thus, Case 1 is the equivalent of the following:
How many ways can 15 chocolates be distributed among 3 children x, y and z so that each child receives at least one chocolate?
To guarantee that each child receives at least 1 chocolate, first give 1 chocolate to each child.
Now count the number of ways to distribute the remaining 12 chocolates among the 3 children.
Since there are 12 chocolates and 3 children, count the number of ways to arrange 12 identical chocolates and 2 identical separators:
14!/(12!2!) =
91.
Since every positive option for x implies a corresponding negative option, we multiply by
2.
Since every positive option for y implies a corresponding negative option, we multiply by another
2.
Since every positive option for z implies a corresponding negative option, we multiply by another
2.
Multplying the values in red, we get:
91*2*2*2 = 728.
Case 2: One integer = 0, the other 2 integers are nonzero
The integer equal to 0 could be x, y or z, implying
3 options.
Here, a sum of 15 is to be distributed among the remaining 2 integers, both of which must be nonzero.
Thus, Case 2 is the equivalent of the following:
How many ways can 15 chocolates be distributed among 2 children so that each child receives at least one chocolate?
To guarantee that each child receives at least 1 chocolate, first give 1 chocolate to each child.
Now count the number of ways to distribute the remaining 13 chocolates between the 2 children.
Since there are 13 chocolates and 2 children, count the number of ways to arrange 13 identical chocolates and 1 separator:
14!/(13!*1) =
14.
Since every positive option for the 1st integer implies a corresponding negative option, we multiply by
2.
Since every positive option for the 2nd integer implies a corresponding negative option, we multiply by another
2.
Multiplying the values in red, we get:
3*14*2*2 = 168.
Case 3: One integer is nonzero, the other two integers = 0
The nonzero integer could be x, y, or z, implying
3 options.
To yield a sum of 15, the nonzero integer must be 15 or -15, implying
2 options.
Multiplying the values in red, we get:
3*2 = 6.
Total number of integral solutions = 728+168+6 = 902.
The correct answer is
A.
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