Evaluate 10 C 7 and 9 P 3?

₁₀C₇
= 10! / [(10 - 7)! 7!]
= 10! / (3! 7!)
= 10 × 9 × 8 / (3 × 2 × 1)
= 120

₉P₃
= 9! / (9 - 3)!
= 9! / 6!
= 9 × 8 × 7
= 504

PERMUTATION:

Let's start with the permutation:
9P3

Start writing down 9 and then multiply by the next smaller number. Stop when you have 3 numbers:
9P3 = 9 * 8 * 7 = 504


COMBINATION:

Now let's do the combination:
10C7

Write it as a fraction:
10
---
7

Now write the next smaller number on top and bottom. Stop when you get to 7 numbers (where it is 1 on the bottom):
10 * 9 * 8 * 7 * 6 * 5 * 4
------------------------------
7 * 6 * 5 * 4 * 3 * 2 * 1

Now simplify by canceling the denominator. Start by crossing out 7 * 6 * 5 * 4
10 * 9 * 8
------------
3 * 2 * 1

Now cancel a 3 from 9:
10 * 3 * 8
--------------
1 * 2 * 1

And cancel a 2 from either even number. I'll pick 8:
10 * 3 * 4
------------
1 * 1 * 1

So what is left is:
10 * 3 * 4
= 120

Answer:
10C7 = 120
9P3 = 504

C(10,7) = 10!/(7!3!) = 10*9*8/(3*2) = 720/6 = 120
P(9,3) = 9!/6! = 9*8*7 = 504