Perms & Combs Halp! (1 Viewer)

[ohexploitable]

I'm so shit at these mang.

20. Find how many ways two numbers can be selected from the numbers 1, 2, ... , 8, 9 so that their sum is:

(c) divisible by 3, (d) divisible by 5, (e) divisible by 6.

22. Nine players are to be divided into two teams of four and one umpire.

(a) In how many ways can the teams be formed?
(b) If two particular people cannot be on the same team, how many different combinations are possible?

23. Four adults are standing in a room that has five exits. Each adult is equally likely to leave the room through any one of the five exits.

(d) What is the probability that no more than two adults come out the same exit?

There will be more to come :-(

 

[hscmathstutor10]

22 is relatively standard ( and there was a very similar one in this years 4unit hsc )

from the group of 9 we pick 4, then from the remaining 5 we pick 4, and the one left over is the umpire

so 22
a. 9C4 x 5C4

b consider the complementary event ie two people on the same team, that means that we kind of ignore 2 people , so we have 7 people left, then from this we pick a group of two ( to add with the 2 already chosen) and then from the remaining 5 we chose 4 , and we multiply the result by 2! as there are two different teams, two different ways for the two people to be on the same team

so ways where they are together = 7 C2 x 5C4 x 2!

therefore, ways where they are not together = ( 9C4 x 5C4 ) - ( 7C2 x 5C4 x 2! )

ill leave the others for someone else to have a go at

 

[ohexploitable]

a. 9C4 x 5C4

Answer says 315 :-(

 

[hscmathstutor10]

they have just divided it by 2 ( or 2! to be more precised )

hhmmm going take a little time to explain

say we had P1 , P2 ... P 9 to stand for the 9 people

well what I did considers these following two events as different ( even though they both end up with the same people in both teams )

( P1 , P2, P3 , P4 ) then ( P5, P6, P7, P8) and P9 as umpire

and ( P5, P6, P7, P8) then ( P1,P2,P3,P4) then P9 as umpire

thats where the extra factor of 2 ( or 2!) comes from

 

[ohexploitable]

I got 9C4*5C4 as well, but I don't know why I overcounted by 2!.

 

[hscmathstutor10]

always better working back from the answer, unforunately you dont get that luxury in the exam

 

[hscmathstutor10]

so then part b should be changed to ( 9C4 x 5C4 / 2! ) - ( 7C2 x 5C4 x 2! / 2! )

hopefully thats correct

 

[ohexploitable]

they have just divided it by 2 ( or 2! to be more precised )

hhmmm going take a little time to explain

say we had P1 , P2 ... P 9 to stand for the 9 people

well what I did considers these following two events as different ( even though they both end up with the same people in both teams )

( P1 , P2, P3 , P4 ) then ( P5, P6, P7, P8) and P9 as umpire

and ( P5, P6, P7, P8) then ( P1,P2,P3,P4) then P9 as umpire

thats where the extra factor of 2 ( or 2!) comes from

Thanks mang.

 

[ohexploitable]

Halp with the others?

 

[hscmathstutor10]

Halp with the others?

I might have a go at 23 if you are lucky, but I dont gaurantee correct answers lol. I remember a very similar hsc question involving rats and mazes but I hardly remember how that one was done

 

[hscmathstutor10]

well Q20 write out the cases

we can get 3, 4, 5, 6, 7,8 9, 10, 11,12,13,14,15,16,17

now ways to be divisable by 3, we can get 3,6,9,12 or 15

ways to get 3 -> (1,2) 1 way
ways to get 6 -> ( 2,4) ( 3,3) (5,1) 3ways
ways to get 9 -> ( 1,8) ( 2,7) (3,6) (4, 5) 4 ways
ways to get 12 -> (1,11) (2,10) (3,9) ( 4,8) ( 5,7) ( 6, 6 ) 6 ways
ways to get 15 -> (1,14) (2,13) (3,12) (4,11) ( 5,10) ( 6,9) ( 7,8) 7 ways

then add them up 21ways in total ( these questions you dont have a choice, you have to write up all the cases )

the other parts for that question are just mind numbly boring

note we are only interested in the combinations, we dont care about the order of selection of the numbers

 

[ohexploitable]

well Q20 write out the cases

we can get 3, 4, 5, 6, 7,8 9, 10, 11,12,13,14,15,16,17

now ways to be divisable by 3, we can get 3,6,9,12 or 15

ways to get 3 -> (1,2) 1 way
ways to get 6 -> ( 2,4) ( 3,3) (5,1) 3ways
ways to get 9 -> ( 1,8) ( 2,7) (3,6) (4, 5) 4 ways
ways to get 12 -> (1,11) (2,10) (3,9) ( 4,8) ( 5,7) ( 6, 6 ) 6 ways
ways to get 15 -> (1,14) (2,13) (3,12) (4,11) ( 5,10) ( 6,9) ( 7,8) 7 ways

then add them up 21ways in total ( these questions you dont have a choice, you have to write up all the cases )

the other parts for that question are just mind numbly boring

note we are only interested in the combinations, we dont care about the order of selection of the numbers

Ewww... I was thinking there had to be an easier way :-(

Anyway, you can't select the same number twice.

Thanks anyway.

 

[deterministic]

23. Four adults are standing in a room that has five exits. Each adult is equally likely to leave the room through any one of the five exits.

(d) What is the probability that no more than two adults come out the same exit?

Consider the complementary case, if more than 2 adults come out the same exit. This means either:
A: 3 adults come out the same exit, one person is different.
B: 4 adults come out the same exit

Note in total there are 5^4=625 ways for 4 adults and 5 doors. Importantly, each way has the same probability of occurring.

Case A:
-choose the 3 adults: 4C3 ways
-choose the doorway for the threesome: 5 ways
-choose the door for the loner: 4 ways
multiply together to get 80 ways.
Prob of this case = 80/total number of ways = 80/625 = 16/125

Case B:
Easy, just choose which door the 4 adults enter = 5 ways
so prob of this case = 5/625 = 1/125

so prob of complementary case = 17/125
then prob of desired event = 1-17/125= 108/125