## Task B-3.2.

How many solutions does the inequality

$$
x_{1}+x_{2}+\ldots+x_{100} \leqslant 103
$$

have for which $x_{1}, x_{2}, \ldots, x_{100}$ are natural numbers?
## Solution.

Since $x_{i}$ are natural numbers, each $x_{i}$ must have a value of at least 1. This means we need to determine in how many ways the sum of natural numbers $x_{1}+x_{2}+\ldots+x_{100}$ can equal 100, 101, 102, and 103.

The sum of 100 can only be obtained in one way, when all addends $x_{i}=1$.

The sum of 101 is obtained if one of the addends is equal to 2, and the other 99 are equal to 1. This addend can be chosen in 100 ways.

The sum of 102 can be obtained in the following cases:

- one addend is equal to 3, and the other 99 are 1 - there are 100 ways for this
- two addends are equal to 2, and the other 98 are 1 - there are $\binom{100}{2}=\frac{1}{2} \cdot 100 \cdot 99$ ways for this.

The sum of 103 can be obtained if:

- one addend is equal to 4, and the other 99 are 1, which is possible in 100 ways
- one addend is equal to 3, one addend is 2, and the other 98 are 1, which is possible in $100 \cdot 99$ ways
- three addends are equal to 2, and the other 97 are 1, which is possible in $\frac{100 \cdot 99 \cdot 98}{3 \cdot 2 \cdot 1}$ ways.

Therefore, the total number of solutions is

$$
1+100+100+\frac{100 \cdot 99}{2}+100+100 \cdot 99+\frac{100 \cdot 99 \cdot 98}{3 \cdot 2 \cdot 1}=176851
$$

solutions to the given inequality.