Question

Find number of natural number solutions for a +b+c+d+e= 20 such that a< b< c<e?​

Answer 1

Given:

a + b + c + d + e = 20.

and, a < b < c < d < e.

To Find:

The number of natural number solutions for the given question.

Solution:

1. Let a = 2, b = 3, c = 4, d = 5 and e = 6.

So,

2 + 3 + 4 + 5 + 6 = 20.

2. Let a = 1, b = 2, c = 4, d = 6 and e = 7.

So,

1 + 2 + 4 + 6 + 7 = 20.

3. Let a = 1, b = 2, c = 3, d = 6 and e = 8.

So,

1 + 2 + 3 + 6 + 8 = 20.

Hence, the above are the three ways of natural number solutions for the given question.