Urgent: Maths CIE Stats

[Deadly_king]

yes thats right thought i have computing tomorrow :/ haha

Hehe.......Good luck then br0

I don't do computing

[$H00t!N& $t@r]

question! 

The following question is from 9709_w09_qp_62:
(i need the solution for 4 a(ii) and b)

4 (a) (i) Find how many different four-digit numbers can be made using only the digits 1, 3, 5 and 6
with no digit being repeated.

(ii) Find how many different odd numbers greater than 500 can be made using some or all of
the digits 1, 3, 5 and 6 with no digit being repeated.

(b) Six cards numbered 1, 2, 3, 4, 5, 6 are arranged randomly in a line. Find the probability that the
cards numbered 4 and 5 are not next to each other.

Thanks in advance 

[lana]

question!  

The following question is from 9709_w09_qp_62:
(i need the solution for 4 a(ii) and b)

4 (a) (i) Find how many different four-digit numbers can be made using only the digits 1, 3, 5 and 6
with no digit being repeated.

(ii) Find how many different odd numbers greater than 500 can be made using some or all of
the digits 1, 3, 5 and 6 with no digit being repeated.

(b) Six cards numbered 1, 2, 3, 4, 5, 6 are arranged randomly in a line. Find the probability that the
cards numbered 4 and 5 are not next to each other.

Thanks in advance  

b) cards numbered 4 and 5 next to each other : 5!× 2!
    6 cards  can be arranged in 6!
    6! - (5!×2!) = 480
    probability : 480/6! = 2/3

[lana]

I have a question !! how can I know which measure of central tendency is the most appropriate to represent a set of data ??!!
thank youu 

[Deadly_king]

I have a question !! how can I know which measure of central tendency is the most appropriate to represent a set of data ??!!
thank youu 

Am not sure myself.

Hope this helps

http://simon.cs.vt.edu/sosci/Site/MMM/mmm.html

http://www.quickmba.com/stats/centralten/

By the way thanks for the answers.
+rep

[lana]

Am not sure myself.

Hope this helps

http://simon.cs.vt.edu/sosci/Site/MMM/mmm.html

http://www.quickmba.com/stats/centralten/

By the way thanks for the answers.
+rep

thank youu !!
and ur so welcome !! =]
thanks for the +rep 

[$!$RatJumper$!$]

Am not sure myself.

Hope this helps

http://simon.cs.vt.edu/sosci/Site/MMM/mmm.html

http://www.quickmba.com/stats/centralten/

By the way thanks for the answers.
+rep

Excellent info!

[lana]

another question !! ( sorry !! )
(just iv and v)
7 Nine cards, each of a different colour, are to be arranged in a line.
(i) How many different arrangements of the 9 cards are possible?
The 9 cards include a pink card and a green card.
(ii) How many different arrangements do not have the pink card next to the green card?
Consider all possible choices of 3 cards from the 9 cards with the 3 cards being arranged in a line.
(iii) How many different arrangements in total of 3 cards are possible?
(iv) How many of the arrangements of 3 cards in part (iii) contain the pink card?
(v) How many of the arrangements of 3 cards in part (iii) do not have the pink card next to the green
card?
thanks =]

[$H00t!N& $t@r]

b) cards numbered 4 and 5 next to each other : 5!× 2!
    6 cards  can be arranged in 6!
    6! - (5!×2!) = 480
    probability : 480/6! = 2/3

Thanks alot! +rep 

[lana]

Thanks alot! +rep 

I'm just glad to help =]
thanks for +rep

[Freaked12]

another question !! ( sorry !! )
(just iv and v)
7 Nine cards, each of a different colour, are to be arranged in a line.
(i) How many different arrangements of the 9 cards are possible?
The 9 cards include a pink card and a green card.
(ii) How many different arrangements do not have the pink card next to the green card?
Consider all possible choices of 3 cards from the 9 cards with the 3 cards being arranged in a line.
(iii) How many different arrangements in total of 3 cards are possible?
(iv) How many of the arrangements of 3 cards in part (iii) contain the pink card?
(v) How many of the arrangements of 3 cards in part (iii) do not have the pink card next to the green
card?
thanks =]

We have to start from part iii to continue with part iv and v
a permutation of a set of values is an arrangement of those values into a particular order.
so 9p3=504

iv)We have to assume that the pink card has been chosen in a group of 3 cards so now that group has 2 places remaining
8P2=56
There will be three groups in total which can be made from a group of 9 cards and 56 times 8 cards other th&n the pink can fit in two places of a group in which the pink card has already been chosen.
so we multiply 56 * 3 =168

[lana]

thank youu soo soo much that was very helpful
+rep =]

[Freaked12]

__ __ 7C1

7C1 __ __

There will be two ways the pink and Green card can shift between themselves as 2!=2
and there are two ways they can stay together

2*2*7=28

504-28=476

[Freaked12]

Explained it in a easier way

~Sf staff at your service

[lana]

Explained it in a easier way

~Sf staff at your service

aaawww you guys are just amazing thank youu =]