Solve 4+4*(4-2+1)+4*(4-1)*(4-3+1)+4!

$$4+4 \times (4-2+1)+4 \times (4-1) \times (4-3+1)+4 !$$

Evaluate

$$64$$

Solution Steps

Subtract 2 from 4 to get 2.

$$4+4\left(2+1\right)+4\left(4-1\right)\left(4-3+1\right)+4!$$

Add 2 and 1 to get 3.

$$4+4\times 3+4\left(4-1\right)\left(4-3+1\right)+4!$$

Multiply 4 and 3 to get 12.

$$4+12+4\left(4-1\right)\left(4-3+1\right)+4!$$

Add 4 and 12 to get 16.

$$16+4\left(4-1\right)\left(4-3+1\right)+4!$$

Subtract 1 from 4 to get 3.

$$16+4\times 3\left(4-3+1\right)+4!$$

Multiply 4 and 3 to get 12.

$$16+12\left(4-3+1\right)+4!$$

Subtract 3 from 4 to get 1.

$$16+12\left(1+1\right)+4!$$

Add 1 and 1 to get 2.

$$16+12\times 2+4!$$

Multiply 12 and 2 to get 24.

$$16+24+4!$$

Add 16 and 24 to get 40.

$$40+4!$$

The factorial of 4 is 24.

$$40+24$$

Add 40 and 24 to get 64.

$$64$$

Factor

$$2^{6}$$