• Best of luck to the class of 2024 for their HSC exams. You got this!
    Let us know your thoughts on the HSC exams here
  • YOU can help the next generation of students in the community!
    Share your trial papers and notes on our Notes & Resources page
MedVision ad

need some help with permuatations and combinations please! (1 Viewer)

physician

Some things never change.
Joined
Oct 21, 2004
Messages
1,432
Location
Bankstown bro
Gender
Male
HSC
2005
ok I need some help with the following 2 questions...

perms. and combs. isn't one of my strong points... so these questions may seem relatively easy to some ppl... anyways....

Q1) In a railway car, with 3 seats facing the engine and 3 seats with their back to the engine, in how many ways can 6 people be seated if two of them insist on facing the engine?

Q2) If an ape typed the letters S, E, E and M in any order what is the probability that he would type the word SEEM?

I pritty much answered the rest of the excersice but got stuck at these two... i would really appreciate ur help...

oh and just incase anyone asks... its from the 3 unit mathematics S.B.Jones, K.E.Couchman blue book excercise 30.4 questoions 11 and 12
 

Alexluby

ZOOM ZOOM ZOOM
Joined
Mar 2, 2005
Messages
28
Location
BAM BAM BAM
Gender
Male
HSC
N/A
The first answer should be 2*1*4!=48

The second should be 4!\2!=12...

Tell me if you want a more detailed answer.
 

Trev

stix
Joined
Jun 28, 2004
Messages
2,037
Location
Pine Palace, St. Lucia, Brisbane.
Gender
Male
HSC
2005
Alexluby said:
The second should be 4!\2!=12...
Actually, 12 is the number of ways the letters S-E-E-M can be arranged, the question asked for the probability of the primate choosing the word SEEM.
So in that case, it would be 1/[1/4*1/3*1/2*1/1*2 ] or 1/[4!/2!] = 1/12. Yes?
 

Pace_T

Active Member
Joined
Oct 21, 2004
Messages
1,784
Gender
Male
HSC
2005
Alexluby said:
The first answer should be 2*1*4!=48


incorrect.

its 3P2*4P4

= 144
which is what is in the textbook.
 

physician

Some things never change.
Joined
Oct 21, 2004
Messages
1,432
Location
Bankstown bro
Gender
Male
HSC
2005
Trev said:
Actually, 12 is the number of ways the letters S-E-E-M can be arranged, the question asked for the probability of the primate choosing the word SEEM.
So in that case, it would be 1/[1/4*1/3*1/2*1/1*2 ] or 1/[4!/2!] = 1/12. Yes?
correct.. yep that's the answer

thanks Trev

yeh ummm could u just explan why its 4!/2! and not 4!/3!...?
 
Last edited:

Trev

stix
Joined
Jun 28, 2004
Messages
2,037
Location
Pine Palace, St. Lucia, Brisbane.
Gender
Male
HSC
2005
physician said:
correct.. yep that's the answer

thanks Trev

yeh ummm could u just explan why its 4!/2! and not 4!/3!...?
Because the letter E appears twice in the word SEEM. Therefore 2 Factorial.
Another example:
HIPPOPOTAMUS
P appears 3 times; O appears twice, so the number of arrangements is 12!/(3!2!).
 

physician

Some things never change.
Joined
Oct 21, 2004
Messages
1,432
Location
Bankstown bro
Gender
Male
HSC
2005
Trev said:
Because the letter E appears twice in the word SEEM. Therefore 2 Factorial.
Another example:
HIPPOPOTAMUS
P appears 3 times; O appears twice, so the number of arrangements is 12!/(3!2!).
oh yeh lol... thankls dude.. i was mixing it up by placing the number of letters available for some odd reason.. i.e. s, e, m... = 3....

lol... thanks i really apprecaite ur help...
 

Trev

stix
Joined
Jun 28, 2004
Messages
2,037
Location
Pine Palace, St. Lucia, Brisbane.
Gender
Male
HSC
2005
Question 1.

O O X; OR; X O X; OR; X X O
Engine
O O O

O are positions that can be changed, X are the ones which want to face the engine.
I get 2*4!*3 = 144 positions.

EDITED: touché PaceT lol.
 
Last edited:

Pace_T

Active Member
Joined
Oct 21, 2004
Messages
1,784
Gender
Male
HSC
2005
physician said:
I don't get it!.. could u just explain it a little...

sure thing dude

ok, there's 3 special seats that 2 particular people must occupy

example:
2 seats that 2 particular people must occupy would be 2P2 (or 2!)

instead its 3 possible seats, and we need to find how many arrangements are possible for these 2 ppl.
hence 3P2 (note that 2P2 = 2! in the example)

now, theres 4 seats left. any of the 4 remaining can take any seat they want. there are no restrictions.
hence its multiplied by 4! (same as 4P4)

therefore its 3P2 * 4P4
 

Pace_T

Active Member
Joined
Oct 21, 2004
Messages
1,784
Gender
Male
HSC
2005
Trev

That is incorrect, you are assuming those two people want two particular seats at the front, when in fact they want any two of the three seats available.
 

Trev

stix
Joined
Jun 28, 2004
Messages
2,037
Location
Pine Palace, St. Lucia, Brisbane.
Gender
Male
HSC
2005
Pace_T said:
Trev

That is incorrect, you are assuming those two people want two particular seats at the front, when in fact they want any two of the three seats available.
Yeah, I didn't think about that :eek: My post is edited. Thanx Pace_T.
 

LaCe

chillin, killin, illin
Joined
Jan 29, 2005
Messages
433
Location
Where am I?
Gender
Undisclosed
HSC
2005
Ok these are pretty easy

all of u still havent answered the second one properly, it asks for the probablilty

therefore it is 1/12 because u have 1/(4!/2!)

the first question is also easy

u have two sets of three seats
well u first place the two ppl, then u can place the remaining seat in 4 ways, then the other three seats 3! ways and then arrange the set of seats facing the engine with the tow ppl on it 3! ways

then u get 1.1.4.3!.3! = 144
 

ngai

Member
Joined
Mar 24, 2004
Messages
223
Gender
Male
HSC
2004
Q1) In a railway car, with 3 seats facing the engine and 3 seats with their back to the engine, in how many ways can 6 people be seated if two of them insist on facing the engine?
deal with the troublemakers first
3 choices for 1st troublemaker (there are 3 seats facing engine)
2 choices for the other (2 seats left)
sort out the other ppl: 4!
so 3*2*4!
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top