a+b+c=10
\(\frac{a+b+c}{3}\) = \(3.33\)
The highest among them can take 7 values(4, 5, 6, 7, 8, 9 and 10)
Case 1- The highest number is 10.
Possible triplet (a,b,c) = (10,0,0)
Total arrangements possible \(= \frac{3!}{2!} = 3\)
Case 2- The highest number is 9.
Possible triplet (a,b,c) = (9,1,0)
Total arrangements possible \(= 3! = 6\)
Case 3- The highest number is 8.
Possible triplets (a,b,c) = (8,1,1) and (8,2,0)
Total arrangements possible \(= \frac{3!}{2!}+3! = 9\)
Case 4- The highest number is 7.
Possible triplets (a,b,c) = (7,2,1) and (7,3,0)
Total arrangements possible \(= 3!+3! = 12\)
Case 5- The highest number is 6.
Possible triplets (a,b,c) = (6,4,0), (6,3,1) and (6,2,2)
Total arrangements possible \(=3!+ 3! +\frac{ 3!}{2!} = 15\)
Case 6- The highest number is 5.
Possible triplets (a,b,c) =(5,5,0), (5,4,1) and (5,3,2)
Total arrangements possible \(= \frac{ 3!}{2!}+3!+3! = 15\)
Case 7- 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 non-negative integers, are there such 'a + b + c = 10' =3+6+9+12+15+15+6 = 66
Bunuel wrote:
How many ordered triplets (a, b, c), where a, b, and c are non-negative integers, are there such a + b + c = 10 ?
A. 54
B. 56
C. 66
D. 72
E. 78
Project PS Butler
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