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16 Sep 2014, 00:46
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D
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Difficulty:
25%
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Question Stats:
82% (01:29) correct
18%
(01:41)
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If \(x\), \(y\), and \(z\) are positive integers, how many different ordered triplets \((x, y, z)\) are possible such that \(x + y + z = 5\)?
A. 3
B. 4
C. 5
D. 6
E. 8
A. 3
B. 4
C. 5
D. 6
E. 8
16 Sep 2014, 00:46
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Official Solution:
If \(x\), \(y\), and \(z\) are positive integers, how many different ordered triplets \((x, y, z)\) are possible such that \(x + y + z = 5\)?
A. 3
B. 4
C. 5
D. 6
E. 8
There are two ways to break the number 5 into the sum of three positive integers: \(2+2+1\) and \(3+1+1\). Since we are interested in ordered triplets, each combination can be represented by three distinct triplets: \((2, 2, 1)\), \((2, 1, 2)\), \((1, 2, 2)\) and \((3, 1, 1)\), \((1, 3, 1)\), \((1, 1, 3)\), respectively.
Answer: D
If \(x\), \(y\), and \(z\) are positive integers, how many different ordered triplets \((x, y, z)\) are possible such that \(x + y + z = 5\)?
A. 3
B. 4
C. 5
D. 6
E. 8
There are two ways to break the number 5 into the sum of three positive integers: \(2+2+1\) and \(3+1+1\). Since we are interested in ordered triplets, each combination can be represented by three distinct triplets: \((2, 2, 1)\), \((2, 1, 2)\), \((1, 2, 2)\) and \((3, 1, 1)\), \((1, 3, 1)\), \((1, 1, 3)\), respectively.
Answer: D
19 Jul 2016, 07:06
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Can this problem be solved using the combinations formula?
