Question

math help ss below

Answer 1

Convert 11.42424242 … to a rational expression in the form of a over b, where b ≠ 0.

Answer 2

i think you posted on the wrong question?

Answer 3

i dont really know how to use this i jus need help on pre algebra

Answer 4

hello?

Answer 5

If you need help post ur own question

Answer 6

The question is asking "how many number of ways..." can this type of hand of cards be dealt. Well, we can walk through it manually to start, and maybe we'll start to notice a pattern.
There are 13 different card "ranks" in total.
We know that our hand will have 2 pairs (of different card ranks), and then the 5th card can be any of the remaining 11 ranks.
We'll have a first pair, a second pair, and a lone card. The first pair can be any of the 13 ranks. The second pair can be any of the remaining 12 ranks. And the lone card can be any of the remaining 11 ranks.
In general, that should mean that there are 13 * 12 * 11 ways to make that combination.

I think, probability is not my strong suite.

Answer 7

hmm i believe that i wrong because the fact that the order of cards doesn't matter so one would have tofind a way to get repeats out?

Answer 8

so would I divide that by 13?

Answer 9

Oh good catch. You're right, with my method there would be repeats.
Why divide by 13? I mean, that might be correct, I just don't understand.

Answer 10

I wasn't sure it's just that i encountered a kinda similar problem and they divided by the base numbers

Answer 11

Maybe I should try to work it out manually.
Pairs of 2 can match with pairs 3 through Ace. So that's 12 pair combinations there.
Then pair 3 can match with 4 through Ace. That's another 11
Pair 4 will give us another 10, pair 5 will give us another 9... Now the pattern is becoming clear.
The pair-groups have 12+11+10+9+8+7+6+5+4+3+2+1 combinations. Then we can take that sum and multiply it by 11, to account for the lone card.
And *that* should give us the total number of combinations.

Answer 12

I think that would be the original method I gave, divided by 2? I think that's what we had to do.

Answer 13

ah that makes sense let me try that out!

Answer 14

oh it was wrong?

Answer 15

I mean what I said before 13*12*11 was wrong.
But this solution I just came up with should be correct, it's actually equal to (13*12*11)/2
Unless it's wrong again, in which case we can think about it some more.

Answer 16

yea it was wrong again thats what i mean

Answer 17

Shoot. Alright, back to the drawing board I guess. Hmmmmm

Answer 18

Hold on, are we supposed to account for the suite of the card as well, like club, diamond, heart, spade? Because that would increase the different possibilities by a *lot*.

Answer 19

i don't think so because it doesn't specfiy the suits in the example

Answer 20

Ok that's good. And we know the order doesn't matter, so we don't need to consider that...
Just curious, what number did you get before from those calculations we tried?

Answer 21

1716 and 858

Answer 22

Ok that's what I got too. Too bad it wasn't the answer. I wonder what it is we're missing here?

Answer 23

maybe the overcounting part?

Answer 24

we're dividing the wrong thing maybe?

Answer 25

If this is probability shouldn't we be doing 2/50 etc.?

Answer 26

Since we aren't putting cards back we also have to change the b of the probability fraction since the source is growing smaller

Answer 27

Aka, 4/52 * 3/52 * 4/52 *3/52 *4/44

Answer 28

The last one is 4/44 because we subtract the 2 ranks we have already pulled.

Answer 29

Oh hold on, the question is asking about "how many different ways can you be dealt 5 cards..." so we'd be looking for a whole number of combinations, not the fractional chance.

It's a probability question, but not directly about probabilities

Answer 30

Huh, good point.

Answer 31

So really we should be looking at how many ranks there are

Answer 32

Which is 13

Answer 33

The chance of getting a card from one rank is 1/13

Answer 34

Then the next one would be 1/12 since it can't be the same rank. And the last would be 1/11

Answer 35

You would do the first two twice.

Answer 36

Try this number:
56628

Answer 37

ah wait they are counting suits! my bad! but i found the answer already thanks for the help however

Answer 38

Suits? As in king queen?

Answer 39

answer if ur curios

Answer 40

I am very confused. Why are those numbers floating?

Answer 41

Thank you! That's actually quite informative, and not as complicated as I assumed

Answer 42

Are they being divided?

Answer 43

Huh! Never heard of that! Thanks for the lesson @smokeybrown

Answer 44

Of course! I'm glad we can all learn together :)