Let XA, XB, XC and XD represent number of balls present in box A, B, C and D respectively.
As no box can be empty so,
XA $$\ge$$ 1, XB $$\ge$$ 1, XC $$\ge$$ 1 and XD $$\ge$$ 1
$$\Rightarrow$$ XA $$-$$ 1 $$\ge$$ 0, $$\Rightarrow$$ XB $$-$$ 1 $$\ge$$ 0, $$\Rightarrow$$ XC $$-$$ 1 $$\ge$$ 0 and $$\Rightarrow$$ XD $$-$$ 1 $$\ge$$ 0
tA $$\ge$$ 0, tB $$\ge$$ 0, tC $$\ge$$ 0 and tD $$\ge$$ 0
According to the question,
XA + XB + XC + XD = 10
$$\Rightarrow$$ (XA $$-$$ 1) + (XB $$-$$ 1) + (XC $$-$$ 1) + (XD $$-$$ 1) = 6
$$\Rightarrow$$ tA + tB + tC + tD = 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 $$ = ({}^3{C_2}) \times ({}^9{C_2}) = 3 \times {{9 \times 8} \over 2} = 108$$