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Jordan Said:

Friend help? (i will appreciate it!!)?

1. do u like her?
2. Is she trust worthy?
3. Do u have fun when ur around her and like being around her?
If u answered yes to these then stay friends with her
My family is richer then a lot of my friends and my parents are more easy going.. And my friends respect me for who I am... So should ur friend... I tink she just wants u to like her... She might even be lying about all those things... Just to get u to like her... I used to use my clothes and dads money to buy my friends... And it DID NOT WORK.. And I'm not the kind of person who would do that.. But I'm still friends with the ppl I tried to buy but we had to earn eachothers trust back.. There's a good chance she's doing that

When she's acting all like that and talking about her money and blah blah... Ignore her... She is showin off and if u ignore her when she says this stuff then after a while she'll get the point that u don't like it
Or u could just flat out tell her "shut up ur bugging me" LOL well maybe not like that!!!

I hate to admit that I used to be just like ur friend (but hey I was in like 4th grade!!!) and well actually I wasn't as showy offy as ur friend but u get the point.... And my family actually really isn't as rich as I thought we were... But were still up there lol!

Anyway hope I helped sweetie!!! And good luck!!!!

Louis Said:

permutations?

Assuming I'm interpreting the worksheet correctly, question 1 asks how many permutations and how many combinations of two vowels you can get out of those five letters.

Well, there are only two vowels, so there's only one such combination. But there are two permutations, because if you change the order ("permute" them), it's a different permutation. Those are:
EA and
AE

Question 2 gets into the fun part. You want both the number of different permutations and the number of combinations possible with three letters, out of the five given.

Let's do permutations first, and let's take the general case: we have N items and want to use k of them. Then the first one of them has N possible choices, but whatever it is, that leaves only (N-1) choices for the next one. That choice, in turn, leaves (N-2) choices for the next one, and (if we were choosing more than 3) so on.

So the general formula for the number of permutations is
N * (N-1) * .... * (N - k + 1)

It's convenient to use "factorial" notation for this, where
N factorial is written N! and
N! = N * (N-1) * (N-2) * ... * 2 * 1

Just in case it comes up, zero factorial = 1 (because it's 1 factorial divided by 1).

But our general formula doesn't always use the smaller multipliers--although it would if we wanted a permutation that used all N items--so we divide by another factorial to cancel out the multipliers we don't use. The number of permutations of k items out of N is therefore
N! / (N-k)!
and where N is 5 and k is 3, that works out to
5! / 2! = 5 * 4 * 3 = 60.

On to combinations: we're making the same selection, but we no longer care about order. But the number of combinations is just the number of permutations, divided by the number of permutations per combination (because each possible combination generates the same number of permutations).

The number of permutations per combination of k items is just k!. Well, actually, using our permutations formula, it's
k! / (k - k)! = k! / 0! = k! / 1 = k!
(which is where that definition of 0! comes in handy).

So the general formula for combinations of k out of N items (commonly called NCk) is
N! / [(N-k)! k!]
and with N=5 and k=3 that's
5C3 = 5! / (2! * 3!)
which with a little clever canceling out becomes
5 * 4 / (2 * 1) = 10.

The 10 three-letter combinations from the set P, E, A, R, S are:
PEA, PER, PAR, EAR, PES, PAS, EAS, PRS, ERS, ARS

Incidentally, you'll notice that because N - (N - k) = k, the general formula for the number of combinations gives the same result whether we choose k items or (N - k) items. That should seem logical, because each way to choose k items to include is also a way to choose (N - k) items to exclude.

That also makes it easier to list combinations sometimes. I listed the 10 combinations in our example by choosing the different pairs of letters to exclude, because it means managing only two choices at a time instead of three.

Let's also look at the first combination and generate the permutations, just as a check:
PEA, PAE, EPA, EAP, APE, AEP
And since every one of the 10 combinations can generate six permutations in the same fashion, the 60 permutations we calculated above is correct.

I realize I've written a lot, but I hope it helps to explain what's behind the formulas.

Next day:
It's occurred to me that I could have been more helpful by including specific examples, so I've gone back and done so.

Eddie Said:

I'm in 7th grade and this year I have a very boring Social Studies teacher?