Let XA, X_B, X_C and X_D represent number of balls present in box A, B, C and D respectively.

As no box can be empty so,

X_A $$\ge$$ 1, X_B $$\ge$$ 1, X_C $$\ge$$ 1 and X_D $$\ge$$ 1

$$\Rightarrow$$ X_A $$-$$ 1 $$\ge$$ 0, $$\Rightarrow$$ X_B $$-$$ 1 $$\ge$$ 0, $$\Rightarrow$$ X_C $$-$$ 1 $$\ge$$ 0 and $$\Rightarrow$$ X_D $$-$$ 1 $$\ge$$ 0

t_A $$\ge$$ 0, t_B $$\ge$$ 0, t_C $$\ge$$ 0 and t_D $$\ge$$ 0

According to the question,

X_A + X_B + X_C + X_D = 10

$$\Rightarrow$$ (X_A $$-$$ 1) + (X_B $$-$$ 1) + (X_C $$-$$ 1) + (X_D $$-$$ 1) = 6

$$\Rightarrow$$ t_A + t_B + t_C + t_D = 6

Now question becomes, box A, B, C, and D can have none or one or more balls and total balls are 6

From formula we know, n things can be distributed among r people in $${}^{n + r - 1}{C_{r - 1}}$$ ways where each people can have either 0 or more things.

$$\therefore$$ 6 balls can be distributed among 4 boxes in $${}^{6 + 4 - 1}{C_{4 - 1}} = {}^9{C_3}$$ ways where each box can have either 0 or more balls.

Therefore, Statement 1 is correct. The number of ways of choosing any 3 places from 9 different places is $${}^9{C_3}$$ ways. But Statement - 2 is not the correct explanation of Statement - 1.

We need 3 points to create a triangle. With 10 points number of triangle possible $${}^{10}{C_3}$$

Here 6 points are on the same line so we can't make any triangle with those 6 points.

So subtract $${}^{6}{C_3}$$.

$$\therefore$$ $$N = {}^{10}{C_3} - {}^6{C_3}$$

$$ = {{10\,.\,9\,.\,8} \over {1\,.\,2\,.\,3}} - {{6\,.\,5\,.\,4} \over {1\,.\,2\,.\,3}}$$

$$ = 120 - 20$$

$$ = 100$$

Thus number of ways $$ = ({}^3{C_2}) \times ({}^9{C_2}) = 3 \times {{9 \times 8} \over 2} = 108$$