Equal to 20

[noreply]

If x, y and z are all positive integers, how many different solutions are there for the equation x + y + z = 20?

[orishorjo]

If x, y and z are all positive integers, how many different solutions are there for the equation x + y + z = 20?
10+5+5=20
15+3+2=20
16+2+2=20
And many more

[ahmed]

1+1+18=20

2+1+17=20

2+2+16=20

3+2+15=20

3+3+14=20

4+3+13=20

4+4+12=20

5+4+11=20

5+5+10=20

6+5+9=20

6+6+8=20

7+6+7=20

7+7+6=20

8+7+5=20

8+8+4=20

9+8+3=20

9+9+2=20

10+9+1=20

1 think there is 18 solutions:cool:

[MAS1]

If x, y and z are all positive integers, how many different solutions are there for the equation x + y + z = 20?

x = 1
---------
y = 1, z = 18
y = 2, z = 17
y = 3, z = 16
.
.
.
y = 18, z = 1

So there are 18 solutions when x = 1.

x = 2
---------
y = 1, z = 17
y = 2, z = 16
y = 3, z = 15
.
.
.
y = 17, z = 1

So there are 17 solutions when x = 2.

If we keep on going the same way for different values of x then there are

18 + 17 + 16 + ... + 2 + 1 = 171 solutions.

[noreply]

Good work guys! Unfortunately, it's more than guessing and checking - and there are 171 solutions! Correct, MAS1!

There's another way too, using combinations.
Notice that 1 + 1 + 1...... + 1 = 20 contains 19 '+' signs.

By choosing to mark two of these plus signs, we divide the left side of the equation (the 1+1+1+1 part) into three parts - one for x, one for y and one for z.

The number of choices we can get is the number of solutions we require. So, the answer is 19C2 (you can put this onto a scientific calculator), which is equal to 171.