Hi! I have tried to figure out these problems so many times and each time my brain almost explodes. The book doesn't explain what (!) is, it just shows a high answer and I'm completely lost. So any help would be appreciated. Thanks!

1) The first problem is "How many different permutations can you make with the letters in the word s e v e n t e e n?" So far I've gotten to : 9! / (4! x 2!). not knowing what (!) is has me stuck.

2) In how many ways can you fill three different positions by choosing from 15 different people? (This question is for confirmation of my answer, 15 x 14 x 13)

3) A teacher has a set of 12 problems to use on a math exam. The teacher makes different versions of the exam by putting 10 questions on each exam. How many different exams can the teacher make?

4) Car thieves steal one automobile out of 400 of a certain type every year in a certain city. What annual net premium should an owner pay for theft insurance in the amount of $16,000 on the certain type of car?(Another confirmation for my answer,$400)

5) Suppose you draw a card from a well-shuffled pack of playing cards. What is the probability the card you draw will be an ace? (And once more another confirmation for my answer, 1/ 52)

6) In how many ways can you arrange eight books on a shelf?

Thanks so much again! If this is too much can you just explain the first few to me so I can figure out the rest. Thanks!

2. Originally Posted by The_Hot_Chick

1) The first problem is "How many different permutations can you make with the letters in the word s e v e n t e e n?" So far I've gotten to : 9! / (4! x 2!). not knowing what (!) is has me stuck.
If all the letters were different we can choose the letters in sequence, so the
first letter can be chosen in 9 ways, the second in 8 ways, the third in 7, ..

So there would be $\displaystyle 9.8.7.6.5.4.3.2.1$ permutations which we write as $\displaystyle 9!$ for short which is read as nine factorial.

Now we have four "e"s and two "n"s. So for each of the $\displaystyle 9!$ permutations in we in fact have $\displaystyle 4!$ repeats with just the "e"s switched around, and $\displaystyle 2!$ with the "n"s switched around, so the total number of distinct permutations is $\displaystyle 9!/(4!.2!)$.

RonL

3. ## touching up

you have equation $\displaystyle \frac{9!}{4!2!}$ here's how to solve...

$\displaystyle \frac{9!}{4!2!}=\frac{9\times8\times7\times6\times 5\times\not{4}\times\not{3}\times\not{2}\times\not {1}}{\not{4}\times\not{3}\times\not{2}\times\not{1 }\times2\times1}$$\displaystyle =\frac{9\times(4\times\not{2})\times7\times6\times 5}{\not{2}}=9\times7\times6\times5\times4$

4. ## I don't have much time...

5) Suppose you draw a card from a well-shuffled pack of playing cards. What is the probability the card you draw will be an ace? (And once more another confirmation for my answer, 1/ 52)
no, there are 4 aces in a deck, therefore 4/52 is the probability, which can be simplified to 1/13

(I will edit more in in a couple hours if noone else answers)

~$\displaystyle Q\!u\!i\!c\!k$

5. Originally Posted by The_Hot_Chick
6) In how many ways can you arrange eight books on a shelf?
For first book you have 8
For second you have 7
For third you have 6
....
Thus,
$\displaystyle 8\times 7\times .... \times 1=8!$

6. Originally Posted by The_Hot_Chick
2) In how many ways can you fill three different positions by choosing from 15 different people? (This question is for confirmation of my answer, 15 x 14 x 13)
It depends what you are asking. If you are asking how many ways there are to select 3 people from 15 then it is,
$\displaystyle _{15}C_3=\frac{15\times 14\times 13}{3\times 2\times 1}$
And if you mean such as there is a first place second place thrid place, such as (1 leader, 1 janitor, 1 teacher) then you have it correct.

7. Originally Posted by The_Hot_Chick
3) A teacher has a set of 12 problems to use on a math exam. The teacher makes different versions of the exam by putting 10 questions on each exam. How many different exams can the teacher make?
This question is like the last question I did. To select 10 from 12.
Which is,
$\displaystyle _{12}C_{10}$
Simplify it as,
$\displaystyle _{12}C_{12-10}=_{12}C_2=\frac{12\times 11}{2!}=66$

8. Originally Posted by ThePerfectHacker
It depends what you are asking. If you are asking how many ways there are to select 3 people from 15 then it is,
$\displaystyle _{15}C_3=\frac{15\times 14\times 13}{3\times 2\times 1}$
And if you mean such as there is a first place second place thrid place, such as (1 leader, 1 janitor, 1 teacher) then you have it correct.
Wouldn't this be $\displaystyle \frac{15!}{3!}$

9. Originally Posted by Quick
Wouldn't this be $\displaystyle \frac{15!}{3!}$
No,
$\displaystyle _nC_m=\frac{n!}{m!(n-m)!}$

Because you are selecting 3 thus it is,
$\displaystyle 15\times 14\times 13$
but then you need to exclude order cuz it is not important here. There are a total of 6 different order arrangement because there are 3 which gives $\displaystyle 6=3!$

10. I get it now..so 8! would mean 8 x 7 x 6...and so forth. So if I multiply all those together then I would get my answer. And with the deck question..all I had to figure out was how many aces in a deck and how many cards in a deck and that is 4/52 then simplify to 1/13. Ahhh okk. Thanks a bunch. Just made math ALOT easier.

11. Hello, The_Hot_Chick!

Here's some help . . .

2) In how many ways can you fill three different positions
by choosing from 15 different people?
(This question is for confirmation of my answer, 15 x 14 x 13)

4) Car thieves steal one automobile out of 400 of a certain type every year in a certain city. What annual net premium should an owner pay for theft insurance
in the amount of $16,000 on the certain type of car? (Another confirmation for my answer,$400)

The problem is worded strangely . . . in fact, sloppy!

should charge for a $16,000 auto-theft policy (in order to "break even") ? [If you ask the owner, "How much should you pay for a$16,000 policy?",
Since $\displaystyle \frac{1}{400}$ of the cars (of that type) are stolen annually,
. . the company expects to pay out $16,000$\displaystyle \frac{1}{400}$of the time. The other$\displaystyle \frac{399}{400}$of the time, they get to keep the premium$\displaystyle (P).$Their "Expected Value" is: .$\displaystyle E\;=\;\frac{1}{400}(-\$16,000) + \frac{399}{400}(P)$
To "break even", the Expected Value is zero: .$\displaystyle -40 + \frac{399}{400}P \:=\:0$
. . and we have: .$\displaystyle P \:= \:\frac{16,000}{399} \:= \:40.10025063$
To break even, the company must charge at least $40.11 per year. ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ Note If they round down to$40.10, they will lose an average of 10 cents per year