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

physician

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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

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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

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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

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Alexluby said:
The first answer should be 2*1*4!=48


incorrect.

its 3P2*4P4

= 144
which is what is in the textbook.
 

physician

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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!...?
 
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Trev

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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

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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

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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.
 
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Pace_T

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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

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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

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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

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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

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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!
 

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