Math combination problem

III You must explain each answer below, not just give a number.

You have four boxes which are all different from each other. One box is red (the others are not).
a. In how many ways can you arrange the four boxes on a shelf?

Would the answer be: 4!=4*3*2*1=24 since you are arranging all 4 boxes

b. In how many ways can you arrange the four boxes so that the first one is not the red box?

Would the answer be: 4c3 = 4/1 = 4 since you are choosing every box but the red box to be first.

c. In how many ways can you arrange the four boxes so that the first one is the red box?

Would the answer be: 4c1 = 4/1 = 4 since you are only choosing the red box to be first

d. In how many ways can you arrange the four boxes so that the red box is not either the first box or the last box?

I am not sure how to answer this one.

e. Assume that you add another box, so there are a total of five boxes, all different from each other. In how many ways can you arrange exactly four of these boxes on the shelf?

Would this be: 5! = 120 since you are arranging all 5 boxes

a. correct
b. You can arrange the 3 boxes in 3! ways and in each of those ways the red box can be put in one of 3 positions: 2nd, 3rd or 4th

3*3! = 18

c. You can arrange the 3 boxes in 3! ways and put the red box 1st.

1*3! = 6

Another way to look at this is that if there are 24 total arrangements and 18 with the red box not 1st (part b), then there must be 24-18=6 with the red box first.

d. You can arrange 3 boxes in 3! ways and put the red box in one of 2 positions: 2nd or 3rd.

2*3! = 12

e. correct

umm 3?