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08 May 2020, 02:03
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Difficulty:
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62% (02:02) correct
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How many ordered triplets (a, b, c), where a, b, and c are positive integers, are there such a + b + c = 10 ?
A. 28
B. 36
C. 45
D. 54
E. 63
Are You Up For the Challenge: 700 Level Questions
08 May 2020, 02:36
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a+b+c=10
\(\frac{a+b+c}{3}\) = \(3.33\)
The highest among them can take 5 values(4, 5, 6, 7 and 8)
Case 1- The highest number is 8.
Possible triplet (a,b,c) = (8,1,1)
Total arrangements possible \(= \frac{3!}{2!} = 3\)
Case 2- The highest number is 7.
Possible triplet (a,b,c) = (7,2,1)
Total arrangements possible \(= 3! = 6\)
Case 3- The highest number is 6.
Possible triplets (a,b,c) = (6,3,1) and (6,2,2)
Total arrangements possible \(= 3! +\frac{ 3!}{2!} = 9\)
Case 4- The highest number is 5.
Possible triplets (a,b,c) = (5,4,1) and (5,3,2)
Total arrangements possible \(= 3!+3! = 12\)
Case 5- The highest number is 4.
Possible triplets (a,b,c) = (4,4,2) and (4,3,3)
Total arrangements possible \(= \frac{3!}{2!} + \frac{3!}{2!} = 6\)
Number of ordered triplets (a, b, c), where a, b, and c are positive integers, are there such 'a + b + c = 10' = 3+6+9+12+6 = 36
\(\frac{a+b+c}{3}\) = \(3.33\)
The highest among them can take 5 values(4, 5, 6, 7 and 8)
Case 1- The highest number is 8.
Possible triplet (a,b,c) = (8,1,1)
Total arrangements possible \(= \frac{3!}{2!} = 3\)
Case 2- The highest number is 7.
Possible triplet (a,b,c) = (7,2,1)
Total arrangements possible \(= 3! = 6\)
Case 3- The highest number is 6.
Possible triplets (a,b,c) = (6,3,1) and (6,2,2)
Total arrangements possible \(= 3! +\frac{ 3!}{2!} = 9\)
Case 4- The highest number is 5.
Possible triplets (a,b,c) = (5,4,1) and (5,3,2)
Total arrangements possible \(= 3!+3! = 12\)
Case 5- The highest number is 4.
Possible triplets (a,b,c) = (4,4,2) and (4,3,3)
Total arrangements possible \(= \frac{3!}{2!} + \frac{3!}{2!} = 6\)
Number of ordered triplets (a, b, c), where a, b, and c are positive integers, are there such 'a + b + c = 10' = 3+6+9+12+6 = 36
