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Re: If x, y and z are non-negative integers such that x < y < z,
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07 Nov 2018, 06:44
3
GreenlightTestPrep 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
Re: If x, y and z are non-negative integers such that x < y < z,
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20 Sep 2021, 06:04
1
Sirish123 wrote:
is there a more quicker version to solve this question? i dont think we will have much time during the actual Exam to list out all the numbers?
There are situations in which counting all of the possible outcomes is actually the fastest approach. Be sure to include the strategy in your mathematical toolbox.
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