Expansion of a large binomial

Expansion. $(1+2x)^7=\sum_i\binom{7}{i}2^ix^i$, $(1-x)^5=\sum_j\binom{5}{j}(-1)^jx^j$.

Coefficient of $x^3$. $\sum_{i+j=3}\binom{7}{i}2^i\binom{5}{j}(-1)^j$.

Compute.
- $i=0,j=3$: $1\cdot1\cdot10\cdot(-1)=-10$
- $i=1,j=2$: $7\cdot2\cdot10\cdot1=140$
- $i=2,j=1$: $21\cdot4\cdot5\cdot(-1)=-420$
- $i=3,j=0$: $35\cdot8\cdot1=280$

Sum: $-10+140-420+280=-10$.

Conclusion. The coefficient is $-10$.