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The calculation 9!/3!6! is equivalent to 9C3 (or 9C6)
Explain clearly why you could solve this question using combinations, and why this is equivalent to considering permutations with repeated items.

- anonymous

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- schrodinger

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- anonymous

@ganeshie8

- anonymous

@nincompoop

- anonymous

Oh thanks!

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## More answers

- anonymous

Thanks for your previous explanations! They were easy to understand:)

- anonymous

What can I do to get an overview of this question?

- ganeshie8

Hey is that the complete problem ?

- anonymous

We have looked at situations in which we need to determine the number of possible routes between two places. We can look at the situation below as 9 steps, six of which must be East and three of which must be South.
This gives us 9!/3!6! = 84 possible routes

- anonymous

##### 1 Attachment

- anonymous

Those are the starting places:)

- anonymous

Thanks for pointing them out

- anonymous

The file "imagen" is exactly as it is

- ganeshie8

|dw:1437035983074:dw|

- anonymous

What may be the discernible connection here?

- ganeshie8

Lets work this problem using "permutations" only first

- anonymous

Ok

- UsukiDoll

maybe we need to calculate the 9!/3!6! first?

- anonymous

6 east and 3 south.

- ganeshie8

A valid route is 9 steps long.
Must have 6 East and 3 South, yes ?

- anonymous

Yes. They have to be the shortest.

- UsukiDoll

I got 6 east 3 south too

- ganeshie8

Like :
\[EEEEEE~SSS\]

- anonymous

9P9?

- ganeshie8

How many distinct words can you make by rearranging the characters in that string above ?

- anonymous

9!/(6!*3!)

- anonymous

Because of the indistinguishable letters

- ganeshie8

How did you get that, could you please explain a bit more..

- UsukiDoll

wasn't that 9!/(6!)(3!) from the original problem?

- anonymous

Because there are 9 letters to be arranged, but permutation 9! would have to be reduced by letters that are identical to each other. And 6! accounts for Es and 3! accounts for Ss. Therefore the total number of distinct words are divided by overlapping ones.

- anonymous

I had a similar question before where I had to rearrange letters to make distinct words, and one of the math tutors here showed me that I need to divide the original all possible options by the factorials of overlapping ones.

- ganeshie8

Perfect!

- ganeshie8

so that is a method using permutations only, lets look at working the total number of routes using "combinations" only

- anonymous

For example, in "mathematics"
there are 2 "m"s
2 identical "a"s
2 identical "t"s
So
11!/(2!*2!*2!)=possible rearranged distinct words from mathematics.

- anonymous

Ok.

- UsukiDoll

\[\frac{n!}{r!(n!-r!)}\] is the formula for the number of possible combinations of r objects in from the set of n objects
it's usually written in nCr.

- ganeshie8

To go from red point on top left corner to the green point on bottom right corner, you need to take 9 steps. 3 of them South and 6 toward East. So if you `choose 3 steps for South out of total 9 steps`, the remaining 6 steps are forced to be East.
|dw:1437036960970:dw|

- anonymous

3 steps south where it could be anywhere, but 6 steps east would have to be made to get to the destination

- ganeshie8

|dw:1437037029139:dw|

- UsukiDoll

so if we have 9C3 that would mean that we let n = 9 and r=3 for nCr
\[\frac{9!}{3!(9!-3!)}\]

- anonymous

How come it's the 3! down below 9!?

- anonymous

I know that 3 souths and 6 easts. But somehow 6 is ignored here?

- anonymous

I understand Ganeshie's easy explanations, but suddenly formatting them in an equation is quite mind numbing.

- ganeshie8

|dw:1437037261640:dw|

- anonymous

3 steps must be south somewhere, and 6 steps must be east. Ok.

- anonymous

The order in which they are taken does not matter?

- anonymous

So long as the 3 steps to the south and 6 steps to the east are achieved .

- ganeshie8

Let me ask you a related question,
Pretend that you have 9 girls as friends, you want to choose 1 girl to go out with.
How many possibilities are there to pick one girl out of 9 girls ?

- anonymous

9 girls.

- anonymous

because there are 9 girls, from whom to pick one.

- ganeshie8

Right, that means you're rejecting 8 other girls right ?

- anonymous

Yes.

- anonymous

So if I pick a certain path I would be rejecting others.

- ganeshie8

As you can see, this question is not that retarded. This is actually a combination problem.
Number of ways of choosing 1 girl from 9 girls = 9C1 = 9
Here we are not worrying about 8 other girls. They are forced to be rejected automatically.

- UsukiDoll

WOW ganeshie!!! that r word!

- ganeshie8

Haha! @Robert136 just want you see that when you choose 3 steps for South, the remaining 6 other steps are automatically forced to be rejected for South.

- UsukiDoll

true you can only go one way after you go 3 steps south |dw:1437037748310:dw|

- UsukiDoll

|dw:1437037774091:dw|

- ganeshie8

So the math is simple :
Total number of routes = number of ways of choosing 3 steps for South out of 9 total steps

- anonymous

and I put focus on 3 steps for south because moving east is a predetermined act?

- ganeshie8

In the exact same spirits,
Total number of routes = number of ways of choosing 6 steps for East out of 9 total steps

- ganeshie8

Yes, when you choose 3 steps for south, the fate of remaining 6 steps is forced.

- anonymous

9C3

- UsukiDoll

\(\color{blue}{\text{Originally Posted by}}\) @UsukiDoll
so if we have 9C3 that would mean that we let n = 9 and r=3 for nCr
\[\frac{9!}{3!(9!-3!)}\]
\(\color{blue}{\text{End of Quote}}\)

- ganeshie8

Yes, that also equals 9C6.
With the earlier analogy, "selecting 1 girl out of 9 girls" is same as "rejecting the 8 other girls".

- UsukiDoll

\[\frac{9!}{3!(9!-3!)} \rightarrow \frac{9!}{3!(6!)}\]

- anonymous

So the reason why both permutation and combination could both be used is because the circumstances in which the question is asked permits the use of both.

- anonymous

Or that conceptually both combination and permutation are relevant in deriving the answer.

- anonymous

Do I still appear retarded? Are my conclusions way offXD?

- UsukiDoll

similarly suppose we do have 9C6 then n=9 and r =6
\[\frac{9!}{6!(9!-6!)} \rightarrow \frac{9!}{6!(3!)}\]
same thing but different order

- anonymous

I see. But that is a coincidence no? The mysteries of numbers

- UsukiDoll

tbh I haven't went into much depth on it..
permutations <-order matters!
combinations <- order doesn't matter

- anonymous

Like anything with the terms of combination method that is arrived at the same factorials

- UsukiDoll

formula for permutations is similar to the combination except the r! is missing

- anonymous

Usuki you still have me blocked I can't write you testimonials yet

- anonymous

XD

- UsukiDoll

\(\color{#0cbb34}{\text{Originally Posted by}}\) @UsukiDoll
\[\frac{n!}{r!(n!-r!)}\] is the formula for the number of possible combinations of r objects in from the set of n objects
it's usually written in nCr.
\(\color{#0cbb34}{\text{End of Quote}}\)
For permutations
\[\frac{n!}{n!-r!}\]
or
\[\frac{n!}{(n-r)!} \]
written in the form nPr

- UsukiDoll

actually that sounds related.. ummm you can refer to this link @Pci
https://www.mathsisfun.com/combinatorics/combinations-permutations.html

- anonymous

Gracias a todos por comentarios muy entiendibles.

- anonymous

If anyone needs Spanish practice in return:)

- ganeshie8

Lets go back to the earlier string :
\[EEEEEE~SSS\]
Consider the problem of finding number of total 9 letter distinct words by rearranging letters in above string.
How many ways can you choose 3 places for S's ?
|dw:1437038371595:dw|