|
My class started the probability and statistics chapter.. and since we took a test today, we're supposed to learn this section ourselves, but why would you do that when you can just ask TL and they'll give help? (hopefully)
Yes, I do know the fundamental counting principle.
But, I don't understand how I'm supposed to make the equation so I can solve it. I might update the OP with more questions that I'll need help on, so don't go away I need this help!
#1) How many ways can six different books be arranged on a shelf if one of the books is a dictionary and it must be on an end?
I know the answer is 240 from the back of the book, but I have no idea where the got the numbers from.
#2) How many different 5-digit codes are possible using a keypad if the first digit isn't zero and no digits are allowed to repeat.
Having the same problem as above, don't know how to set up the equation. =\
a keypad is
789
456
123
x0x
in case you didn't know. 
Thanks for the help! (If I get any, ._.)
I thought about using TL Manpower, but I don't really have the time to wait for a PM (._. i'm sorry I'm impatient and blogs get faster replies and I also don't know the available times of the people on TL Manpower[Maybe a 'convenient times' could be a part of the TL Manpower resume?] so I decided to go with a blog. Now that I have taken a few minutes out of my time to type this, won't you please take a few minutes out of your time to do some math for me?
 
|
I guessed girl blog, actually
|
1) 5! (for the ways 5 books can be arranged in any order) x 2 (number of places the dictionary can be placed)
5x4x3x2x1x2=240
2) Does the question mean repeat in sequence or repeat as in be used a 2nd time. Ex. 311 or 313 being incorrect?
|
Oohh Niceee explanation of #1
but #2 just means that it has to be like 12345 no numbers can be used more than once
|
United States24653 Posts
Same Question as NovaTheFeared...
If it just means you can't repeat it twice in a row, then the answer should be 9^5 since there are nine options other than zero for the first digit, and then nine options that don't repeat thereafter.
edit: oh in that case it's trickier
first digit: 9 options
second digit: 9 options
third digit: 8 options
fourth: 7 options
fifth: 6 options
9*9*8*7*6
|
Considering the difficulty of the first question I'm going with instinct.
9 possibilities for the first digit (1-9)
9 possibilities for the 2nd digit (0-9, less the one chosen)
8 possibilities for the 3rd digit
7 possibilities for the 4th digit
6 possibilities for the 5th digit
9*9*8*7*6=27216
|
This stuff is easy as hell, what grade are you in? It makes me think you're extremely lazy to not try to learn this stuff yourself.
#1: If dictionary is first then 5x4x3x2x1 = 120 different positions
The numbers are the number of possible books that could be 2nd,3rd,4th,5th,6th
If dictionary is last then also 5x4x3x2x1 = 120 different positions
The numbers are the number of possible books that could be 1st, 2nd, 3rd, 4th, 5th
240 altogether
#2: 9x9x8x7x6 = 27216
The numbers are the number of possible digits that could be 1st, 2nd, 3rd, 4th, 5th
So ridiculously easy -_-
|
ohh dang
i see now!!
i'm just a freshman in high school D=
don't judge me plz
|
Okay uhh next question I'm stuck on (not really stuck I could count this, but I don't know how to setup a formula for this)
How many numbers between 100 & 999 have 7 in the tens place?
Not sure how I would set this up...
would it be like
XYZ
X= hunderds place
Y= Tens place
Z= ones place?
|
it is 9x1x10 = 90. X can be a number from 1-9 so 9 choices. Y can only be 7 so one choice. Z can be a number from 0-9 so 10 choices.
|
Ahhh I see!!
cool i set it up right
I'm gettin the hang of this shitt yeaaaaaaaaaaa
-is proud- 
|
On February 19 2009 10:55 NovaTheFeared wrote:
1) 5! (for the ways 5 books can be arranged in any order) x 2 (number of places the dictionary can be placed)
On February 19 2009 10:59 NovaTheFeared wrote:
Considering the difficulty of the first question I'm going with instinct.
9 possibilities for the first digit (1-9)
9 possibilities for the 2nd digit (0-9, less the one chosen)
8 possibilities for the 3rd digit
7 possibilities for the 4th digit
6 possibilities for the 5th digit
9*9*8*7*6=27216
Your homework is done.
|
On February 19 2009 12:01 vx70GTOJudgexv wrote:Show nested quote +On February 19 2009 10:55 NovaTheFeared wrote:
1) 5! (for the ways 5 books can be arranged in any order) x 2 (number of places the dictionary can be placed) Show nested quote +On February 19 2009 10:59 NovaTheFeared wrote:
Considering the difficulty of the first question I'm going with instinct.
9 possibilities for the first digit (1-9)
9 possibilities for the 2nd digit (0-9, less the one chosen)
8 possibilities for the 3rd digit
7 possibilities for the 4th digit
6 possibilities for the 5th digit
9*9*8*7*6=27216 Your homework is done.
And you are more useless than this post is
|
I did figure out the first question after years without doing stats. As other people have explained for first one, the dictionary has to be the end, then you can only arrange the other 5 books, that leaves 5x4x3x2=120. It says it has to be an end, and you can have it at the front or at the very last, so 120x2 = 240. Didn't attempt #2, have to get back to iccup.
|
United States24653 Posts
On February 19 2009 12:01 OmgIRok wrote:Ahhh I see!! cool i set it up right I'm gettin the hang of this shitt yeaaaaaaaaaaa -is proud-
Be careful. When receiving help its easy to overestimate how much you yourself have actually gotten from the experience. If you took a quiz on this topic tomorrow, how do you think you would do? You'd probably do worse :p
|
On February 19 2009 12:50 micronesia wrote:Show nested quote +On February 19 2009 12:01 OmgIRok wrote:Ahhh I see!! cool i set it up right I'm gettin the hang of this shitt yeaaaaaaaaaaa -is proud- Be careful. When receiving help its easy to overestimate how much you yourself have actually gotten from the experience. If you took a quiz on this topic tomorrow, how do you think you would do? You'd probably do worse :p
Eh.. the fact that I finished the rest of my homework without needing anymore help says a lot. (too me it does)
|
Micronesia brings up a good point, but I'd like to elaborate.
If you didn't understand at all, and this got you started on the right path, then you've done a good thing by asking for help. If you stop here, you've done a bad thing, since the help you recieved won't stick with you. You want to take what you've learned here, and try and apply it yourself in different situations (ie- problems). Once you can produce a correct answer and are able to explain the reasoning behind it without looking at an example or asking for help, you'll be ready to be tested.
|
|
|
|
|
|
EPT
