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Yikes tomorrow is the day! Just going over circled questions in destroyer, chads quizzes, and reading through alan's notes. But I didnt have the time/money to get math destroyer, but ive studied and timed myself with the achiever/regular destroyer questions and am decently comfortable except for permutations. Anyone know a good site for learning them? Chad does a good job explaining them but the questions with multiple probabilities and factorials get confusing. AHHHHHHHHHHHHH.

I hate permutations and all that too.. I heard Khan's Academy videos are decent for math. Good luck man!

clancy THE BEST WAY IS BELOW:

If the order doesn't matter, it is a Combination.

i.e. How many possible ways can 10 people sit in 3 chairs?

Instead of doing a sloppy mess with the n! R! and all that other junk!! do it this way:

(10*9*8) / 3! = 240

The reason why you do 10*9*8 is because thats the highest values going to the lowest and you stop at 8 because you only need 3 chairs, BUT since order does not matter you have to divided by factorial of the # of necessary chairs i.e. 3! in this case.

Try this it works ALL THE TIME!

If the order does matter it is a Permutation.

i.e. Out of 5 people how many can ways can there be Vice President, President, Director (selected in that order)

answer: 5*4*3 = 60

Again you have 5 people, but order matters because you cannot reselect a person to be another position, therefore you take your total 5 and then multiple by 4*3 (again since there are only 3 total positions) but since order MATTERS, you DO NOT divide here!

Thats it! no kaka dudu formulas! this works ALL THE TIME! Try it for yourself, the only time it wont work is if you confuse Order matters for Order doesnt matter; which is easy to do so be careful!

I never do the formulas anymore because this way is logical and less confusing as long as you know whether its a permutation or combination.

thank you !

thumbs up. Thank you!

I always get these probability questions wrong also. seems easier than how answer keys explain

what about this one:

20 chairs are in a row, 4 of the chairs have a broken leg, what is the probability the first four chairs having exactly 1 broken chair leg.

b/c similar questions like this were asked on achiever, which fused probability with permutations/combinations and it got overwhelming.

so much easier now, the formulaes always get me confused

hmm...

Probability for having a non-broken chair = 16/20

Probability for having a broken chair = 4/20

Number of ways to arrange one broken chair + three non-broken chair = 4!/(1!3!) = 4

So (16/20)(16/20)(16/20)(4/20) * 4 = (4/5)^4 = 256/625

I don't know if that's right or not but I think that's what my answer would be.

A formula used here:

"The number of permutations possible out of n objects, (where p of one kind are alike, q of another kind are alike, ... and r of yet another kind are like) can be expressed as:

n!/(p! q!...r!) where p + q + ... + r = n "

Hope this helps.

If you use wierd's method then shouldn't it be

20 x 19 x 18 x 17 = 116280 ?

Im not sure if this is right because we are arranging them in order so no dividing by anything right

Doing that way isn't the correct way mainly because that's the number of permutations that you can arrange 4 chairs out of 20 different chairs (different = order matters), where 20P4 happens.

My answer takes into account that there were two different kinds of chairs, broken and non-broken. If I didn't use permutation to express it, I could still easily say that there were 4 different ways to arrange one broken and three non-broken chairs

* = broken, ~= non-broken

1. *~~~

2. ~*~~

3. ~~*~

4. ~~~*

Notice that in one kind of chair, the order doesn't matter because they are chairs of the same kind. But between the kinds of chairs, order does matter, and that's why you use the formula I posted above.

And you need to take into account of the probabilities of each event happening. So...

1. *~~~ = (4/20)(16/20)(16/20)(16/20)

2. ~*~~ = (16/20)(4/20)(16/20)(16/20)

3. ~~*~ = (16/20)(16/20)(4/20)(16/20)

4. ~~~* = (16/20)(16/20)(16/20)(4/20)

Adding up all of these probabilities (since they are all probable events), give you the answer I posted above.

Hope it helped.

the formula i wrote out only works for simple combination/permutations. when its pretty simple and quick. the only that clancy wrote takes more thought and might take one or two formulas and manipulating to get it right.

You'll do great tomorrow!!!! GOOD LUCK!!!!!!!

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