Answer:
Step-by-step explanation:
Given that there are three variables satisfying the equation
[tex]X1 + x2 + x3 = 22[/tex]
Here each x is given to be a positive integer
i.e. solution set for each of the variable can be any integer from 1 to 20 at most.(because if two other integers are 1 each third has to be 20)
Hence solution set can be of the form
[tex](x1,x2,x3) =(1,1,22) (1,2,21) (1,3,20).....[/tex]
[tex]=(2,1,19) (2,2,18),...\\=(3,1,18) (3,2,17),....\\...\\...\\=(20,1,1)[/tex]
If x1 =1, there are 20 solution sets
If x1 =2,there are 19
...
If x1 =20 there is 1 set
Hence total solutions can be[tex]= 20+19+...+1\\=\frac{20(21)}{2} =210[/tex]
Answer:
210
Step-by-step explanation:
Consider 22 counters ("stars") placed in a row. There are 21 spaces between them. The problem here amounts to finding the number of ways that 2 separators ("bars") can be put in those spaces, dividing the row into 3 parts.
The number of ways 21 objects can be chosen 2 at a time is ...
21C2 = 21!/(2!(21-2)!) = 21·20/(2·1) = 210
There are 210 positive integer solutions to the given equation.
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Here are the first few solutions, to give you the idea:
* | * | * * * ... ⇒ (x1, x2, x3) = (1, 1, 20)
* | * * | * * ... ⇒ (x1, x2, x3) = (1, 2, 19)
* | * * * | * ... ⇒ (x1, x2, x3) = (1, 3, 18)
