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longgone's avatar

Will you help me understand this maths problem?

Asked by longgone (19608points) November 4th, 2015

I’m taking a maths course this semester. Last week, I missed a lesson. I’m catching up, but my professor briefly described a particular problem today, and I’m having trouble with it. Here’s what she wrote:

10! / 3! x 7!

The answer, apparently, is 120.

I’ve done some research, and I know that 3! = 1×2 x 3. That’s about it, though. Help, please?

This could be called a homework question. I’m not asking for the answer, though, I’m just asking for pointers – which is fine, per the guidelines.

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11 Answers

citizenearth's avatar

You already knew the principles of factorial. So you should be having no problem solving this homework question. First, take 10! divided by 7!, that lefts (10×9x8) to be divided by (3×2x1) resulting in answer of 120. Hope you find it easy.

JLeslie's avatar

^^Why 10! Divided by 7!?

LuckyGuy's avatar

You know that factorial numbers get huge quickly . So often there is a trick to the problem
Let’s say you saw a problem like this one: 25! / 24! You could spend the time and try to do the math to figure out what 25! is (hint : a jillion) and divide it by 24! (hint a gazillion) OR you can use this trick:
In your mind write out the factors 25! /24! = 1×2x3×4x5…x23×24x25 / 1×2x3…x23×24 . Cancel out the factors that are on top and on the bottom (all the numbers from 1 to 24) leaving 25 as the answer. Got it?
Now look at 25! /23! The answer is 24×25 = 600 .
Try 15! / 13! Factor out the 13 numbers they have in common to get 14×15 = 210
Let’s toss in an extra step 15!/(13! x 2) That is 14×15 /2 = 105.

The trick is to imagine writing it out in your head (or on scratch paper) and then cancel. If you find yourself calculating huge numbers, STOP. You are doing it wrong.

Extra credit. Which is larger a jillion or a gazillion? How many times lager is it?

LostInParadise's avatar

Continuing with what @LuckyGuy said, there is a second level of simplification. For positive integers n and k, n!/[k!(n-k)!] is always a whole number. After you do the suggested cancelling, you are left with 10*9*8/(3*2*1). Cancel the 3 with the 9 and the 2 with the 5, leaving 5*3*8 = 120.

From your example, it looks like you may be studying combinatorics. I have a Web page on introductory combinatorics. Arthur Benjamin’s book, Magic of Math, which you previously mentioned having, has a nice section on combinatorics also. We both use poker hands as examples. Great minds run in the same channels (or all fools think alike).

JLeslie's avatar

Now I get it. :)

LuckyGuy's avatar

@LostInParadise I wanted to write my answer in equation form but I figured an example would make it easier to visualize and understand.
10!/8!? Cancel out the 1 through 8 leaving 9×10 = 90! Bamm!
200!/199! = ... 200

Nice web page!

longgone's avatar

Wow, I had no idea it was this simple. I got it, so much so that I don’t even have any questions. Many thanks!!

LuckyGuy's avatar

@longgone Great! Did you wonder where factorials are used? Usually in probability theory.

I have 4 chips of different colors. If I draw them out of a box one at a time and place them in a row on a table how many different color patterns can I get? . Let’s see the first time I pull one out there are 4 choices. then next time there are 3 choices, the next time 2 choices ,and finally 1 left. so 4×3x2×1 4! = 24

Let’s do the same problem with 5 jelly beans of 5 different flavors. 5×4x3×2x1 = 5! = 0 Because, seriously, you and I know you will eat them before they hit the table.

dxs's avatar

Remember your order of operations. Is it (10!) / (3! x 7!) or is it (10! / 3!) x 7! ? The latter is how you wrote it, which gives you a huge answer that isn’t 120. The former does. Here’s the problem in fraction form simplified—the numerator (10!) and the denominator (3! x 7!):

10×9 x 8 x 7×6 x 5×4 x 3×2 x 1
3×2 x 1 x 7×6 x 5×4 x 3×2 x 1


10×9 x 8 = 720 = 120
3×2 x 1 ____6

Did that help? (Ignore the underscores in the last line, that’s just for spacing.)

CWOTUS's avatar

I would also recommend the Wolfram Alpha website to help explain and graph all kinds of math problems and equations, and also to assist with well-written definitions.

longgone's avatar

Thanks again! I’m falling in love with this, and I’m bookmarking all your links! :]

@dxs Yeah, I had trouble with the formatting and gave up eventually. Good point, though!

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