If x, y and z can only take the values 1, 2, 3, 4, 5, 6, 7 then find the number of solutions of the equation x + y + z= 12.
Full solution with explaination
Answers
Case 1 : when x,y,z are distinct
(x,y,z) = (1,4,7) is a possible arrangement satisfying x+y+z=12.
(1,4,7) can be arranged among themselves in 3! = 6 ways. Thus we have 6 solutions.
[For example, arrangements of 1,4,7 gives the following 6 solutions satisfying x+y+z=12.
(x,y,z) = (1,4,7), (1,7,4), (4,1,7), (4,7,1), (7,1,4), (7,4,1) ]
Similarly, each of the following values of (x,y,z) also gives 6 solutions.
(1,5,6), (2,3,7), (2,4,6), (3,4,5)
i.e., another 4*6 = 24 solutions
In case 1, total solutions = 6+24 = 30
Case 2 : any two variables are same
Its easy to list down all such arrangements
(3,3,6), (3,6,3), (6,3,3),
(5,5,2), (5,2,5), (2,5,5)
In case 2, total solutions = 6
Case 3 : all variables are same
Only one arrangement is possible here which is (4,4,4)
In case 3, total solutions = 1
Total number of solutions = 30 + 6 + 1= 37
Answer:
Total number of solutions = 30 + 6 + 1= 37
Step-by-step explanation:
Case 1 : when x,y,z are distinct
(x,y,z) = (1,4,7) is a possible arrangement satisfying x+y+z=12.
(1,4,7) can be arranged among themselves in 3! = 6 ways. Thus we have 6 solutions.
For example, arrangements of 1,4,7 give the following 6 solutions satisfying x+y+z=12.
(x,y,z) = (1,4,7),(4,7,1), (1,7,4), (4,1,7), (7,1,4), (7,4,1)
Similarly, the following values of (x,y,z) also gives 6 solutions each.
(1,5,6), (2,3,7), (2,4,6), (3,4,5)
Total number of solutions from (1,4,7)(1,5,6), (2,3,7), (2,4,6), (3,4,5)
5*6 = 30 solutions
Case 2: when two variables are the same
It's easy to list down all such arrangements
(3,3,6), (3,6,3), (6,3,3),
(5,5,2), (5,2,5), (2,5,5)
In case 2, total solutions = 6
Case 3: all variables are the same
Only one arrangement is possible here which is (4,4,4)
Number solutions in this case = 1
Therefore total number of solutions = 30 + 6 + 1= 37.
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