Formula to calculate the sum of combinations

Question

I want the formula to calculate the sum of following pattern given n and m:

$$^nC_1 * ^mC_1$$ $$+^nC_1 * ^mC_2$$ $$...$$ $$+^nC_1 * ^mC_m$$
$$+^nC_2 * ^mC_1$$ $$+^nC_2 * ^mC_2$$ $$...$$ $$+^nC_2 * ^mC_m$$
$$.$$ $$.$$ $$.$$
$$+^nC_n * ^mC_1$$ $$+^nC_n * ^mC_2$$ $$...$$ $$+^nC_n * ^mC_m$$

Thus if n=2 and m=3, I want the formula to calculate the sum of all these:

$$^2C_1 * ^3C_1$$ $$+^2C_1 * ^3C_2$$ $$+^2C_1 * ^3C_3$$
$$+^2C_2 * ^3C_1$$ $$+^2C_2 * ^3C_2$$ $$+^2C_2 * ^3C_3$$

$$=21$$

if n=3 and m=3, then the sum of all these nine items:

$$^3C_1 * ^3C_1$$ $$+^3C_1 * ^3C_2$$ $$+^3C_1 * ^3C_3$$
$$+^3C_2 * ^3C_1$$ $$+^3C_2 * ^3C_2$$ $$+^3C_2 * ^3C_3$$
$$+^3C_3 * ^3C_1$$ $$+^3C_3 * ^3C_2$$ $$+^3C_3 * ^3C_3$$

$$=49$$

So, what can be a general formula through which I can directly get the answer for this pattern given n and m values.

Answer 1

Your sums of products can be broken up into the product of two sums, each of the form $\sum_{k=1}^{m} \binom {m}{k} =2^m-1$, so the answer to your question is $(2^m-1)(2^n-1)$.