If x, y and z are non-negative integers such that x < y &

Brent@GMATPrepNow wrote:If x, y and z are non-negative integers such that x < y < z, then the equation x + y + z = 11 has how many distinct solutions?

A) 5
B) 10
C) 11
D) 22
E) 78

I created this question to show that there can be times when the best (i.e., fastest) way to solve a counting question is by listing and counting

How do we know when it's not a bad idea to use listing and counting?
The answer choices will tell us (ALWAYS scan the answer choices before beginning any answer choices)
Here, the answer choices are reasonably small, so listing and counting shouldn't take long.
ASIDE: Yes, 78 (answer choice E) is pretty big. However, if you start listing possible outcomes and you eventually list more than 22 outcomes (answer choice D), you can stop because the answer must be E.

Okay, let's list possible outcomes in a systematic way.
We'll list outcomes as follows: x, y, z

Since x is the smallest value.
Let's list the outcomes in which x = 0. We get:
0, 1, 10
0, 2, 9
0, 3, 8
0, 4, 7
0, 5, 6

Now, the outcomes in which x = 1. We get:
1, 2, 8
1, 3, 7
1, 4, 6

Now, the outcomes in which x = 2. We get:
2, 3, 6
2, 4, 5

Now, the outcomes in which x = 3. We get:
3, 4...hmmm, this won't work.

So, we're done listing!
Count the outcomes to get a total of 10

Answer: B

Cheers,
Brent

Brent@GMATPrepNow wrote:If x, y and z are non-negative integers such that x < y < z, then the equation x + y + z = 11 has how many distinct solutions?

A) 5
B) 10
C) 11
D) 22
E) 78

Since 11/3 < 4, so x, being the smallest of all 3 variables, must be no more than 3.

If x = 0, then we have y + z = 11. Since y < z, y can be any integers from 1 to 5 inclusive. So we have 5 distinct solutions when x = 0.

If x = 1, then we have y + z = 10. Since y < z, y can be any integers from 2 to 4 inclusive. So we have 3 distinct solutions when x = 1.

If x = 2, then we have y + z = 9. Since y < z, y can be any integers from 3 to 4 inclusive. So we have 2 distinct solutions when x = 2.

If x = 3, then we have y + z = 8. Since y > x, y is at least 4. If y ≥ 4, then z ≤ 4. However, this will contradict the fact that y < z. So we don’t have any possible solutions when x = 3.

Therefore, there are a total of 5 + 3 + 3 = 10 distinct solutions.

Answer: B