anonymous
  • anonymous
MEDAL+FAN+TESTIMONIAL 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.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@ganeshie8
anonymous
  • anonymous
@nincompoop
anonymous
  • anonymous
Oh thanks!

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Thanks for your previous explanations! They were easy to understand:)
anonymous
  • anonymous
What can I do to get an overview of this question?
ganeshie8
  • ganeshie8
Hey is that the complete problem ?
anonymous
  • 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
  • anonymous
1 Attachment
anonymous
  • anonymous
Those are the starting places:)
anonymous
  • anonymous
Thanks for pointing them out
anonymous
  • anonymous
The file "imagen" is exactly as it is
ganeshie8
  • ganeshie8
|dw:1437035983074:dw|
anonymous
  • anonymous
What may be the discernible connection here?
ganeshie8
  • ganeshie8
Lets work this problem using "permutations" only first
anonymous
  • anonymous
Ok
UsukiDoll
  • UsukiDoll
maybe we need to calculate the 9!/3!6! first?
anonymous
  • anonymous
6 east and 3 south.
ganeshie8
  • ganeshie8
A valid route is 9 steps long. Must have 6 East and 3 South, yes ?
anonymous
  • anonymous
Yes. They have to be the shortest.
UsukiDoll
  • UsukiDoll
I got 6 east 3 south too
ganeshie8
  • ganeshie8
Like : \[EEEEEE~SSS\]
anonymous
  • anonymous
9P9?
ganeshie8
  • ganeshie8
How many distinct words can you make by rearranging the characters in that string above ?
anonymous
  • anonymous
9!/(6!*3!)
anonymous
  • anonymous
Because of the indistinguishable letters
ganeshie8
  • ganeshie8
How did you get that, could you please explain a bit more..
UsukiDoll
  • UsukiDoll
wasn't that 9!/(6!)(3!) from the original problem?
anonymous
  • 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
  • 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
  • ganeshie8
Perfect!
ganeshie8
  • ganeshie8
so that is a method using permutations only, lets look at working the total number of routes using "combinations" only
anonymous
  • 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
  • anonymous
Ok.
UsukiDoll
  • 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
  • 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
  • anonymous
3 steps south where it could be anywhere, but 6 steps east would have to be made to get to the destination
ganeshie8
  • ganeshie8
|dw:1437037029139:dw|
UsukiDoll
  • 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
  • anonymous
How come it's the 3! down below 9!?
anonymous
  • anonymous
I know that 3 souths and 6 easts. But somehow 6 is ignored here?
anonymous
  • anonymous
I understand Ganeshie's easy explanations, but suddenly formatting them in an equation is quite mind numbing.
ganeshie8
  • ganeshie8
|dw:1437037261640:dw|
anonymous
  • anonymous
3 steps must be south somewhere, and 6 steps must be east. Ok.
anonymous
  • anonymous
The order in which they are taken does not matter?
anonymous
  • anonymous
So long as the 3 steps to the south and 6 steps to the east are achieved .
ganeshie8
  • 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
  • anonymous
9 girls.
anonymous
  • anonymous
because there are 9 girls, from whom to pick one.
ganeshie8
  • ganeshie8
Right, that means you're rejecting 8 other girls right ?
anonymous
  • anonymous
Yes.
anonymous
  • anonymous
So if I pick a certain path I would be rejecting others.
ganeshie8
  • 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
  • UsukiDoll
WOW ganeshie!!! that r word!
ganeshie8
  • 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
  • UsukiDoll
true you can only go one way after you go 3 steps south |dw:1437037748310:dw|
UsukiDoll
  • UsukiDoll
|dw:1437037774091:dw|
ganeshie8
  • 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
  • anonymous
and I put focus on 3 steps for south because moving east is a predetermined act?
ganeshie8
  • 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
  • ganeshie8
Yes, when you choose 3 steps for south, the fate of remaining 6 steps is forced.
anonymous
  • anonymous
9C3
UsukiDoll
  • 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
  • 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
  • UsukiDoll
\[\frac{9!}{3!(9!-3!)} \rightarrow \frac{9!}{3!(6!)}\]
anonymous
  • 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
  • anonymous
Or that conceptually both combination and permutation are relevant in deriving the answer.
anonymous
  • anonymous
Do I still appear retarded? Are my conclusions way offXD?
UsukiDoll
  • 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
  • anonymous
I see. But that is a coincidence no? The mysteries of numbers
UsukiDoll
  • UsukiDoll
tbh I haven't went into much depth on it.. permutations <-order matters! combinations <- order doesn't matter
anonymous
  • anonymous
Like anything with the terms of combination method that is arrived at the same factorials
UsukiDoll
  • UsukiDoll
formula for permutations is similar to the combination except the r! is missing
anonymous
  • anonymous
Usuki you still have me blocked I can't write you testimonials yet
anonymous
  • anonymous
XD
UsukiDoll
  • 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
  • UsukiDoll
actually that sounds related.. ummm you can refer to this link @Pci https://www.mathsisfun.com/combinatorics/combinations-permutations.html
anonymous
  • anonymous
Gracias a todos por comentarios muy entiendibles.
anonymous
  • anonymous
If anyone needs Spanish practice in return:)
ganeshie8
  • 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|