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#2 Re: Help Me ! » Probability Question » 2015-02-25 17:57:46

Heron wrote:

What is the probability that two of the cards dealt were Jacks and one was a Queen?

He wants exactly two Jacks and one Queen to be dealt.

#3 Re: Help Me ! » Probability Question » 2015-02-25 12:55:26

Hej Heron,

What you have calculated is probability of getting JJQ, in that order. You also have JQJ and QJJ, which have the same probabilities as JJQ, so it's just 3*1/7=3/7.

#4 Re: Help Me ! » Help with a fairly simple proof please? » 2015-02-25 11:39:08

bobbym wrote:

The two polynomials 5n+3 , 7n+4 are already factored and obviously have no common factor.

They are not polynomials, they are numbers.

#5 Re: Introductions » Just a quick hello : ] » 2015-02-21 15:46:22

Hej stevesy

Välkommen till forumet! (Welcome to the forum!) smile

Vad är din favoritfärg? (What is your favourite colour?)

#7 Re: Help Me ! » Surface Area Math Question » 2015-02-03 13:54:26

No. Without any further information, this object have any shape and size.

Do you know anything else about the object?

#8 Re: Help Me ! » Last two digits » 2015-02-03 04:04:56

He's an element of the set P.

Also:

When

,
is divisible by 100, so its last two digits are 00. Because of that, the last two digits of
are the same as those of

#10 Re: Computer Math » A multivariable gf » 2015-01-25 14:06:47

I like the idea of one variable tracking which numbers were picked and the other how many of each was taken.

Not \cdots but \vdots. Right beneath the second line of LaTeX, instead of those three dots.

#11 Re: Computer Math » A multivariable gf » 2015-01-25 13:57:01

Thanks so much for this. I was trying to figure out a GF for this problem, but got nowhere.

By the way, I think \vdots might look better.

#12 Re: Help Me ! » Combinations with repetitions problem » 2015-01-18 09:10:13

Okay, so I think it would go like this:


The number of ways of choosing a combination of flavors is
, because those are combination with replacement. Then we also multiply by 3, for the sizes. So, there are
combinations. From those we need to choose 12 different ones, which we can do in
ways.

#13 Re: Help Me ! » Combinations with repetitions problem » 2015-01-18 06:09:57

So, flavors xi and yi are the same, right?

#14 Re: Help Me ! » Combination Problem » 2015-01-17 15:38:52

Agnishom wrote:

I'm using this principle to solve the second problem:

a. Generate a list of five matches.
b. Map a cartesian product over each set of five matches.
c. Check if the result is <= 500

Won't you just get 25 ordered pairs?

#15 Re: Help Me ! » Combination Problem » 2015-01-17 15:22:32

Well, you never specified that you don't want the two letters that are grouped together to both appear in a combination.

#17 Re: Help Me ! » Combination Problem » 2015-01-17 12:50:14

I think it does. I am getting 8030 possibilities now. I'll wait for ElainaVW to confirm or correct me.

#18 Re: Help Me ! » Combination Problem » 2015-01-17 12:03:48

Hi ElainaVW

Thanks.

Hi NickMeyers

One question - does it matter if J or S was chosen in the last example?

#19 Re: Help Me ! » Combination Problem » 2015-01-17 11:29:04

Hi ElainaVW

Could you hide that?

#20 Re: Help Me ! » Combination Problem » 2015-01-17 10:46:49

The one Elaina mentioned above, Mathematica.

#21 Re: Help Me ! » Combination Problem » 2015-01-17 10:37:12

Hi NickMeyers

Welcome to the forum! smile

There are 6304 combinations for the new list.

#22 Re: Help Me ! » Matrice questions. Please help. » 2015-01-15 13:05:24

For Q1, use the fact that a matrix is non-invertible if its determinant is 0. Do you know how to calculate the determinant of that matrix?

#23 Re: Help Me ! » Infinite Series » 2015-01-11 13:51:41

I'd say no, but I was referring to post #1.

#24 Re: Help Me ! » Infinite Series » 2015-01-11 05:18:46

Both approaches need to be made more rigorous, I think. I should note that those equations are all valid only under the circumstance that |x|<1. You need to prove that the sum converges in order to use that.

#25 Re: Help Me ! » Infinite Series » 2015-01-11 04:53:02

Well, you differentiate term by term.

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