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August 28, 2016
Posted by **Audrey** on Friday, October 19, 2007 at 6:14pm.

a)60 b)100 c)360 d)630 e)720

we're still doing this at school and I still don't get it; since I have to use three of the eight letters I only have five left and can only use two of them, so should I multiply 8 by 5?

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**Anonymous**, Friday, October 19, 2007 at 6:28pmThis problem is a permutation problem. The hint is given in the word "arrangement." However, you could make this to a combination problem by multiplying by r! (since nCr=(nPr)/r! and multiplying it to r! would give you nPr). To do this problem, you have how many can you choose from? (that would be your n) and how many do you choose? (that would be your r)

8 C 3 (8 choose 3)

To enter it in the graphing calculator, you enter 8, cPr, then 3. Changing it into permutation, multiply the result by 3!. - maths (reword) -
**Anonymous**, Friday, October 19, 2007 at 6:52pmwhat is the question asking specifically?

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**Audrey**, Friday, October 19, 2007 at 8:32pmhow many possible combinations there were and thank you very much for your explanation :)

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**Reiny**, Friday, October 19, 2007 at 8:11pmtreat "net" as if it were one item

eg. let X="net"

so now you are simply arranging the X plus 2 of the remaining 5 letters.

Put the X first, then the other two places can be filled in 5x4 ways, which is 20 ways

But the X, the "net" could also be in the middle, or at the end, so there are 3 ways for the X to go

the number of ways is then 3x5x4 = 60

**or**

Since the X (the "net") now makes your letters spell MAGXIC, which is 5 letters, and we have to fill 3places, the number of ways = 5x4x3 = 60- maths -
**Audrey**, Friday, October 19, 2007 at 8:32pmoh, okay, wow, so that's the metohd; I get it now; thanks!

before I had no idea how to even set up the equation, but I'm strating to see the pattern!

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**Audrey**, Friday, October 19, 2007 at 8:35pmI get where the five is coming from, since that's the remainder of letters, but where are you getting the four from?

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**Reiny**, Friday, October 19, 2007 at 9:12pmlet's pretend we put the X down first, which is really the "net" occupying 3 letter positions

That leaves 2 more positions to be filled from the remaining 5 letters, the MAGIC.

So there are 5 ways to fill the next position, right?

Since we must have picked one of those 5 letters, that would leave 4 letters remaining for the next and last spot.

Hence the 5x4, I multiplied by 3 because the "net" could have been first, in the middle or at the end

Hope that makes sense

e.g. Suppose I want to make all possible arrangements consisting of 4 letters, those 4 letters coming from the word COMPUTER, with no letter appearing more than once

no. of arrangements = 8x7x6x5 = 1680 - one more thing, though -
**Audrey**, Friday, October 19, 2007 at 11:19pmoh, okay!

thanks a million! :)

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