First problem:
How many ways are there to distribute $26$ identical balls into six distinct boxes such that the number of balls in each box is odd?
Let $x_k$ denote the number of balls placed in the $k$th box, $1 \leq k \leq 6$. Then
$$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 26 \tag{1}$$
Since each $x_k$ is a positive odd integer, $x_k = 2y_k + 1$ for some non-negative integer $y_k$. Substituting into equation 1 and simplifying yields
$$y_1 + y_2 + y_3 + y_4 + y_5 + y_6 = 10 \tag{2}$$
which is an equation in the non-negative integers. The number of solutions of equation 2 is the number of ways five addition signs can be inserted into a row of ten ones.
There are $$\binom{10 + 5}{5} = \binom{15}{5}$$ solutions since we must choose which five of the fifteen symbols (ten ones and five addition signs) will be addition signs.
Second problem:
How many ways are there to distribute $26$ identical balls into six distinct boxes such that the first three boxes contain at most six balls?
We wish to solve the equation
$$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 26 \tag{3}$$
in the non-negative integers subject to the constraints that $x_1, x_2, x_3 \leq 6$. Equation 3 is an equation in the non-negative integers. It has
$$\binom{26 + 5}{5} = \binom{31}{5}$$
solutions.
We must exclude those cases in which one or more of the constraints is violated.
Suppose the constraint $x_1 \leq 6$ is violated. Then $x_1 \geq 7$. Let $y_1 = x_1 - 7$. Then $y_1$ is a non-negative integer. Substituting $y_1 + 7$ for $x_1$ in equation 3 and simplifying yields
$$y_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 19 \tag{4}$$
Equation 4 is an equation in the non-negative integers. It has
$$\binom{19 + 5}{5} = \binom{24}{5}$$
solutions. By symmetry, equation 3 has the same number of solutions if the constraint $x_2 \leq 6$ or the constraint $x_3 \leq 6$ is violated. Hence, there are
$$\binom{3}{1}\binom{24}{5}$$
solutions in which one of the constraints is violated.
Suppose the constraints $x_1 \leq 6$ and $x_2 \leq 6$ are both violated. Let $y_1 = x_1 - 7$; let $y_2 = x_2 - 7$. Then $y_1$ and $y_2$ are non-negative integers. Substituting $y_1 + 7$ for $x_1$ and $y_2 + 7$ for $x_2$ and simplifying yields
$$y_1 + y_2 + x_3 + x_4 + x_5 + x_6 = 12 \tag{5}$$
which is an equation in the non-negative integers with
$$\binom{12 + 5}{5} = \binom{17}{5}$$
solutions. By symmetry, equation 3 has the same number of solutions if both the constraints $x_1 \leq 6$ and $x_3 \leq 6$ are violated or both the constraints $x_2 \leq 6$ or $x_3 \leq 6$ are violated. Hence, there are
$$\binom{3}{2}\binom{17}{5}$$
solutions in which two of the constraints are violated.
Suppose that all three constraints are violated. Let $y_k = x_k - 7$, $1 \leq k \leq 3$. Then each $y_k$ is a non-negative integer. Substituting $y_k + 7$ for $x_k$, $1 \leq k \leq 3$, and simplifying yields
$$y_1 + y_2 + y_3 + x_4 + x_5 + x_6 = 5 \tag{6}$$
which is an equation in the non-negative integers with
$$\binom{5 + 5}{5} = \binom{10}{5}$$
solutions.
By the Inclusion-Exclusion Principle, the number of ways the $26$ identical balls can be distributed into six distinct boxes such that the first three boxes contain at most six balls is
$$\binom{31}{5} - \binom{3}{1}\binom{24}{5} + \binom{3}{2}\binom{17}{5} - \binom{3}{3}\binom{10}{5}$$
Are you familiar with stars and bars method?
So we have to find the number of 6-tuples $(x_1,...x_6)$ such that each $x_i$ is even number grater than $0$ and their sum is 100. So we can write each $x_i=2y_i$ and we have $$y_1+y_2+...+y_6 = 50$$
If we write $z_i= y_i-1$ we have $$z_1+z_2+...+z_6 = 44$$ and that we can write on $${44+5\choose 5}$$ ways.
Best Answer
There are $5$ boxes. Put a ball in each. So all that remains to calculate is how to place the $20$ remaining balls in the $5$ boxes.
Can you take it from here?