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Hej Bob

That is correct. (Det är rätt.)

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.

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.

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.

Hej stevesy

Välkommen till forumet! (Welcome to the forum!)

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

Hi ganesh

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

Do you know anything else about the object?

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 ofI 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.

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.

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.

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

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?

Hm, is that Haskell?

Hi ElainaVW

Thanks.

Hi NickMeyers

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

Hi ElainaVW

Could you hide that?

The one Elaina mentioned above, Mathematica.

Hi NickMeyers

Welcome to the forum!

There are 6304 combinations for the new list.

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

Well, you differentiate term by term.