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anonymous
 one year ago
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.
anonymous
 one year ago
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.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks for your previous explanations! They were easy to understand:)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What can I do to get an overview of this question?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Hey is that the complete problem ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0We 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
 one year ago
Best ResponseYou've already chosen the best response.0Those are the starting places:)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks for pointing them out

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The file "imagen" is exactly as it is

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3dw:1437035983074:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What may be the discernible connection here?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Lets work this problem using "permutations" only first

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2maybe we need to calculate the 9!/3!6! first?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3A valid route is 9 steps long. Must have 6 East and 3 South, yes ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes. They have to be the shortest.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2I got 6 east 3 south too

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Like : \[EEEEEE~SSS\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3How many distinct words can you make by rearranging the characters in that string above ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Because of the indistinguishable letters

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3How did you get that, could you please explain a bit more..

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2wasn't that 9!/(6!)(3!) from the original problem?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Because 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
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.3so that is a method using permutations only, lets look at working the total number of routes using "combinations" only

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For 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.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{n!}{r!(n!r!)}\] 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
 one year ago
Best ResponseYou've already chosen the best response.3To 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
 one year ago
Best ResponseYou've already chosen the best response.03 steps south where it could be anywhere, but 6 steps east would have to be made to get to the destination

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3dw:1437037029139:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2so if we have 9C3 that would mean that we let n = 9 and r=3 for nCr \[\frac{9!}{3!(9!3!)}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How come it's the 3! down below 9!?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I know that 3 souths and 6 easts. But somehow 6 is ignored here?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I understand Ganeshie's easy explanations, but suddenly formatting them in an equation is quite mind numbing.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3dw:1437037261640:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.03 steps must be south somewhere, and 6 steps must be east. Ok.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The order in which they are taken does not matter?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So long as the 3 steps to the south and 6 steps to the east are achieved .

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Let 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
 one year ago
Best ResponseYou've already chosen the best response.0because there are 9 girls, from whom to pick one.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Right, that means you're rejecting 8 other girls right ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So if I pick a certain path I would be rejecting others.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3As 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
 one year ago
Best ResponseYou've already chosen the best response.2WOW ganeshie!!! that r word!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Haha! @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
 one year ago
Best ResponseYou've already chosen the best response.2true you can only go one way after you go 3 steps south dw:1437037748310:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2dw:1437037774091:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3So the math is simple : Total number of routes = number of ways of choosing 3 steps for South out of 9 total steps

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and I put focus on 3 steps for south because moving east is a predetermined act?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3In the exact same spirits, Total number of routes = number of ways of choosing 6 steps for East out of 9 total steps

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Yes, when you choose 3 steps for south, the fate of remaining 6 steps is forced.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2\(\color{blue}{\text{Originally Posted by}}\) @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
 one year ago
Best ResponseYou've already chosen the best response.3Yes, that also equals 9C6. With the earlier analogy, "selecting 1 girl out of 9 girls" is same as "rejecting the 8 other girls".

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{9!}{3!(9!3!)} \rightarrow \frac{9!}{3!(6!)}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So 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
 one year ago
Best ResponseYou've already chosen the best response.0Or that conceptually both combination and permutation are relevant in deriving the answer.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Do I still appear retarded? Are my conclusions way offXD?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2similarly 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
 one year ago
Best ResponseYou've already chosen the best response.0I see. But that is a coincidence no? The mysteries of numbers

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2tbh I haven't went into much depth on it.. permutations <order matters! combinations < order doesn't matter

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Like anything with the terms of combination method that is arrived at the same factorials

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2formula for permutations is similar to the combination except the r! is missing

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Usuki you still have me blocked I can't write you testimonials yet

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2\(\color{#0cbb34}{\text{Originally Posted by}}\) @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 \[\frac{n!}{(nr)!} \] written in the form nPr

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2actually that sounds related.. ummm you can refer to this link @Pci https://www.mathsisfun.com/combinatorics/combinationspermutations.html

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Gracias a todos por comentarios muy entiendibles.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If anyone needs Spanish practice in return:)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Lets 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