Part A
Since order matters, we use the nPr permutation formula
We use n = 12 and r = 8
![_{n}P_{r} = \frac{n!}{(n-r)!}\\\\_{12}P_{8} = \frac{12!}{(12-8)!}\\\\_{12}P_{8} = \frac{12!}{4!}\\\\_{12}P_{8} = \frac{12*11*10*9*8*7*6*5*4*3*2*1}{4*3*2*1}\\\\_{12}P_{8} = \frac{479,001,600}{24}\\\\_{12}P_{8} = 19,958,400\\\\](https://tex.z-dn.net/?f=_%7Bn%7DP_%7Br%7D%20%3D%20%5Cfrac%7Bn%21%7D%7B%28n-r%29%21%7D%5C%5C%5C%5C_%7B12%7DP_%7B8%7D%20%3D%20%5Cfrac%7B12%21%7D%7B%2812-8%29%21%7D%5C%5C%5C%5C_%7B12%7DP_%7B8%7D%20%3D%20%5Cfrac%7B12%21%7D%7B4%21%7D%5C%5C%5C%5C_%7B12%7DP_%7B8%7D%20%3D%20%5Cfrac%7B12%2A11%2A10%2A9%2A8%2A7%2A6%2A5%2A4%2A3%2A2%2A1%7D%7B4%2A3%2A2%2A1%7D%5C%5C%5C%5C_%7B12%7DP_%7B8%7D%20%3D%20%5Cfrac%7B479%2C001%2C600%7D%7B24%7D%5C%5C%5C%5C_%7B12%7DP_%7B8%7D%20%3D%2019%2C958%2C400%5C%5C%5C%5C)
There are a little under 20 million different permutations.
<h3>Answer: 19,958,400</h3>
Side note: your teacher may not want you to type in the commas
============================================================
Part B
In this case, order doesn't matter. We could use the nCr combination formula like so.
![_{n}C_{r} = \frac{n!}{r!(n-r)!}\\\\_{12}C_{8} = \frac{12!}{8!(12-8)!}\\\\_{12}C_{8} = \frac{12!}{4!}\\\\_{12}C_{8} = \frac{12*11*10*9*8!}{8!*4!}\\\\_{12}C_{8} = \frac{12*11*10*9}{4!} \ \text{ ... pair of 8! terms cancel}\\\\_{12}C_{8} = \frac{12*11*10*9}{4*3*2*1}\\\\_{12}C_{8} = \frac{11880}{24}\\\\_{12}C_{8} = 495\\\\](https://tex.z-dn.net/?f=_%7Bn%7DC_%7Br%7D%20%3D%20%5Cfrac%7Bn%21%7D%7Br%21%28n-r%29%21%7D%5C%5C%5C%5C_%7B12%7DC_%7B8%7D%20%3D%20%5Cfrac%7B12%21%7D%7B8%21%2812-8%29%21%7D%5C%5C%5C%5C_%7B12%7DC_%7B8%7D%20%3D%20%5Cfrac%7B12%21%7D%7B4%21%7D%5C%5C%5C%5C_%7B12%7DC_%7B8%7D%20%3D%20%5Cfrac%7B12%2A11%2A10%2A9%2A8%21%7D%7B8%21%2A4%21%7D%5C%5C%5C%5C_%7B12%7DC_%7B8%7D%20%3D%20%5Cfrac%7B12%2A11%2A10%2A9%7D%7B4%21%7D%20%5C%20%5Ctext%7B%20...%20pair%20of%208%21%20terms%20cancel%7D%5C%5C%5C%5C_%7B12%7DC_%7B8%7D%20%3D%20%5Cfrac%7B12%2A11%2A10%2A9%7D%7B4%2A3%2A2%2A1%7D%5C%5C%5C%5C_%7B12%7DC_%7B8%7D%20%3D%20%5Cfrac%7B11880%7D%7B24%7D%5C%5C%5C%5C_%7B12%7DC_%7B8%7D%20%3D%20495%5C%5C%5C%5C)
We have a much smaller number compared to last time because order isn't important. Consider a group of 3 people {A,B,C} and this group is identical to {C,B,A}. This idea applies to groups of any number.
-----------------
Another way we can compute the answer is to use the result from part A.
Recall that:
nCr = (nPr)/(r!)
If we know the permutation value, we simply divide by r! to get the combination value. In this case, we divide by r! = 8! = 8*7*6*5*4*3*2*1 = 40,320
So,
![_{n}C_{r} = \frac{_{n}P_{r}}{r!}\\\\_{12}C_{8} = \frac{_{12}P_{8}}{8!}\\\\_{12}C_{8} = \frac{19,958,400}{40,320}\\\\_{12}C_{8} = 495\\\\](https://tex.z-dn.net/?f=_%7Bn%7DC_%7Br%7D%20%3D%20%5Cfrac%7B_%7Bn%7DP_%7Br%7D%7D%7Br%21%7D%5C%5C%5C%5C_%7B12%7DC_%7B8%7D%20%3D%20%5Cfrac%7B_%7B12%7DP_%7B8%7D%7D%7B8%21%7D%5C%5C%5C%5C_%7B12%7DC_%7B8%7D%20%3D%20%5Cfrac%7B19%2C958%2C400%7D%7B40%2C320%7D%5C%5C%5C%5C_%7B12%7DC_%7B8%7D%20%3D%20495%5C%5C%5C%5C)
Not only is this shortcut fairly handy, but it's also interesting to see how the concepts of combinations and permutations connect to one another.
-----------------
<h3>Answer: 495</h3>