# Lab: More Counting Practice (#21)

## Problem: Shades of Pre-registration
### a.
$k\times (k-1)\times (k-2)\times (k-3)\times (k-4)$

Let's say we have a set of courses c = {CSC-151, CSC-161, CSC-207}

Then we have one course which we will select the first round. We have three options in our set. Say we randomly select CSC-151, we no longer want to select CSC-151 because we already have it. Then we would have two courses to chose from and then one the last round. Since order matters, each course placement is unique so for each first option we chose there will always be two options for the next slot and one option for the last slot. This follows our combinatiorial description

### b.
$(k \times (k-1) \times (k-2))/(3\times 2\times 1)$
Let's say we have a set of courses c = {A, B, C, D}
choose A we have B C or D and one of the left overs
ABC
ABD
ACD
Then combinations without A that don't repeat
BCD
So we write it normaly for all the combinations at first, but notice that normally there are 3 options for A to be in A_ _, _ A _, and _ _ A then the same for the subsequent letters there are 2 options and 1 option so we must divide by those options

### c.
$4 \times 3 \times (k-2) \times (k-3)$

Let's say we have a set of courses c = {Basket Weaving, Underwater Knitting, CSC-151, CSC-161, CSC-207}

First choose Basket Weaving and Underwater Knitting positions. There are 4 spots to choose for the first class and 3 positions left to choose for the next one whether that be Basket Weaving or Underwater Knitting. If Basket Weaving and Underwater Knitting take up two spots, we are left with two unselected spots. The first spot can be chosen from k-2 courses and the second spot can be chosen from k-3 courses. For our example k would be 5 because there are 5 courses in our total list. The remaining positions would be created out of the remaining CSC course of which there are tree options for the first of the two oen spots and then two options for the last spot because we would have chosen one of those three.

### d.
$k \times (k-1) \times (k-2) \times (k-3) - 4 \times 3 \times (k-2) \times (k-3)$

Using the same set from c. The number of courses is the total number of courses which as we explained before in the similar problem of a will be $k \times (k-1) \times (k-2) \times (k-3)$ and then subtract the result we got in c to get the total for couses without Basket Weaving and Underwater Knitting.

$k \times (k-1) \times (k-2) \times (k-3)$ gives us the possible schedules we can construct from k courses. If we substract $4 \times 3 \times (k-2) \times (k-3)$ from the the schedules that we got, we have the number of four course schedules that do not contain both Basket Weaving and Underwater Knitting.

### e.
$(5+4+3+2+1) \times (k-2) \times (k-3)\times (k-4)\times (k-5)$
Let's say we have a set of courses c = {Calculus I, Computer Science I, CSC-151, CSC-161, CSC-207}

When Calculus I is the first position, that leaves us with 5 spots to fill in Computer Science I; when Calc is in the second position, that leaves us with 4 spots; when calc is in the third position that leaves us with 3 spots...the same logic applies to the rest of the spots and we get 15 possible schedules with Calc and CS as a result with 4 out of 6 spots left to be filled in the normal manner.

### f.(Bonus Problem)
$(5+4+4+4+4+4)\times (k-2) \times (k-3)\times (k-4)\times (k-5)$
Similarly, part e, when Cal 1 is in the first spot, CS1 can be in 5 different spots. While when Cal 1 is in the second spot, except for the first spot, CS1 can be in 4 different spots. Similarly, for when Cal 1 is the third/fourth/fifth/sixth spot, and CS1 can still be in 4 different spots. Thus, we get (5+4+4+4+4+4) 25 possible result with 4 out of 6 spots left to be filled in the normal manner.