A man can hit a target once in 4 shots. If he fires 4 shots in success

There are various ways of getting the answer. Here are a few of them.

Method 1: (Simplest) To hit the target, you need to hit it at least once in the 4 shots. You would not have hit the target, if you do not hit it in any of the 4 shots.
Probability of not hitting the target in any of the 4 shots = (3/4)*(3/4)*(3/4)*(3/4) = 81/256
Probability of hitting the target at least once = 1 - 81/256 = 175/256

Method 2: You hit the target if you hit it at least once. Say, if you hit the target on the first shot, you are done no matter what you do on the rest of the 3 shots. If you do not hit the target on the first shot, but hit it on the second shot, again, you are done after that... etc
Probability of hitting the target = (1/4) + (3/4)*(1/4) + (3/4)*(3/4)*(1/4) + (3/4)*(3/4)*(3/4)*(1/4) = (1*64 + 3*16 + 9*4 + 27)/256 = 175/256

Method 3: Say H = Hit and M = Miss
Probability of hitting the target once = (1/4)*(3/4)*(3/4)*(3/4) * 4!/3! = 27*4/256
Why do we multiply by 4!/3!? because 1 Hit and 3 Misses can be arranged in 4 ways: HMMM or MHMM or MMHM or MMMH. You arrange 4 things out of which 3 are identical in 4 ways.

Probability of hitting the target twice = (1/4)*(1/4)*(3/4)*(3/4) * 4!/2!*2! = 54/256
Why do we multiply by 4!/2!*2!? Same logic as above. You arrange 4 things out of which 2 pairs are identical in 4!/2!*2! ways.

Probability of hitting the target thrice = (1/4)*(1/4)*(1/4)*(3/4) * 4!/3! = 12/256

Probability of hitting the target four times = (1/4)*(1/4)*(1/4)*(1/4) = 1/256

Add them all up. You will get 175/256