C(n,p) = n! / p! (n-p)!

C(10,3) = 10! / 3! (10 – 3)! =** 10! / 3! (7!)** = **8*9*10 / 1*2*3** = 720/6 = 120

My confusion lies in two aspects: 1. The denominator position of two factorials. 2. The factorial calculation of both denominator and numerator.

1.) What do 3! and 7! represent? Or rather why do they both occupy the same denominator space if they do not interact at all? There is no multiplication, addition, subtraction between 3! and 7!, they simply exist independent of one another, together? And why is it that we only use the 3! when calculating (1*2*3), as opposed to both 3! and 7! (1*2*3 *** **4*5*6*7)? Or even just using the 7! as opposed to 3!, why one over the other?

2.) This question reflects the last question in “1.)”, but now I’m asking as to why we’re only doing the last 3 multiplications up to 10! ?

I thought 10! meant 1*2*3*4*5*6*7*8*9*10, not 8*9*10? I’d like to know how, or why, we are only taking only the last 3 of the 10! ?

I assume it deals with “p” in the C(n,p) formula, but not seeing the connection as to why/how that is taking place.

Hi Bryan,

It does seem confusing at first, especially when learning the process. Removing the 1 through 7 is just cutting out all the extra numbers because multiplication and division cancel each other out. 7! / 7! = 5,040 / 5,040 = 1

10! = 3,628,800

3! (7!) = 6 * 5,040 = 30,240

3,628,800 / 30,240 = 120

Likewise,

10 * 9 * 8 (7 and under are excluded) = 720

3! (again, 7 and under are excluded) = 6

720 / 6 = 120 >>> same answer 😀

1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 = 10!

1* 2 * 3 * 4 * 5 * 6 * 7 = 7!

Both 1 – 7 in the numerators and denominators are not necessary to complete the division problem.

Hope this helps!

Stephen

Thanks Stephen! This helps a great deal. Can I ask 2 follow-up questions? So if I see something like “3! (7!)” –> 3! (7!) = 6 * 5,040 = 30,240, it is safe to assume it is implying that they multiply? The 2 factorials next to one another? I guess I’m usually familiar with them having a ” * ” next to one another to imply multiplication. My last question is comprehension feedback: I can understand why we use the 7! to cancel within the 10! (it’s a bigger factorial than 3!, so its a smoother break), but can we do the inverse? And if not, why not? I assume that equation would go like….C(10,3) = 10! / 3! (7!) = 4*5*6*7*8*9*10 / 4*5*6*7 = 604800 / 840 = 720 … which is our numerator answer for my above posted question…but not over 6 to be equal to 120. So maybe I missed a step or not quite getting the cancellation process, but can you comment on the inverse or rather picking the 3! to cancel as opposed the 7! (aside it logistically being more cogent to pick 7!)? Or if it is even possible?

Ciao, Yes. Anytime you see a parentheses directly next to another variable, it is implied it is multiplication. With factorials also, even 3!7! is implied a multiplication of the two factorials. I guess it’s just for the sake of not writing out so much stuff, I dunno lol. But keep your parentheses in mind. It’s also always the first thing you want to address in an equation. Remember your order of operations through PEMDAS and be careful with how the question is constructed 🙂 e.g.: 3(3^2)+ 4 = 31 but (3(3^2)) + 4 = 85 Q1: Parentheses & Exponent: (3^2) = 9 >> Mult. 3 = 27 >>> Add 4 = 31. In the second one the exponent, “^2” would be distributed to both 3’s, leaving it as 3^2 x 3^2 + 4 >>> 9 x 9 + 4 = 85. The second question, I guess it doesn’t hurt to try it out from the smaller integer end and go up and see if answers match. I personally don’t do that. I’ve learned to do factorials from large no. > small but like I said earlier with the denominator in question, the parentheses – (7!) in this case – would take precedence over 3! and thus not giving you the answer you want. Hope this clarifies some more! Cheers.

Stephen B., MS

Cool beans! That officially clears up my confusion. Thanks for all your help. Bryan Y., OVO