even more permutations! T_T (1 Viewer)

[Smilebuffalo]

Hey guys i fail at this topic. Need some help in figuring out how these questions are done. Much appreciated


23. Find the number of arrangements of the letters in the word pencils if:
i) e precedes i (answer = 2520)
ii) there are 3 letters between e and i (answer = 720)


28. A car can hold 3 people in the front seat and 4 in the back seat. In how many ways can 7 people be seated in the car if 2 particular people must sit in the back seat and 1 particular person is the driver? (answer = 288)

29. In how many ways can 4 people be accommodated if there are 4 rooms available? (answer = 256) <--- how come the answer isn't simply 4P4?

33. In how many ways can 5 men and 5 women be arranged in a circle so that the men are separated? In how many ways can this be done if two particular women must not be next to a particular man? (answer = 2880; 864)

 

[zeleboy]

29. In how many ways can 4 people be accommodated if there are 4 rooms available? (answer = 256) <--- how come the answer isn't simply 4P4?

Each individual has 4 choices. Therefore there are 4^4 choices available. (fitzpatrick ay)

 

[hermand]

23.i.
if e is in the first position, i can be in six positions, and the rest is 5P5.
if e is in the second position, i can be in five positions and the rest is 5P5.
if e is in the third position, i can be in four positions and the rest is 5P5.
etc, so the equation comes out as..

 

[ninetypercent]

28. A car can hold 3 people in the front seat and 4 in the back seat. In how many ways can 7 people be seated in the car if 2 particular people must sit in the back seat and 1 particular person is the driver?

ways in which those 2 people can sit: 4P2 (selecting 2 seats out of 4)
ways in which the other people can sit: 4!
total: 4P2 x 4! = 288

 

[hermand]

23.ii.
working it out similarly to the above question,
if e is in the first position, i can be in three positions and the rest are 5P5.
if e is in the second position, i can be in two positions and the rest are 5P5.
if e is in the third position, i can be in one position and the rest are 5P5.
which makes..

 

[lychnobity]

29. In how many ways can 4 people be accommodated if there are 4 rooms available? (answer = 256) <--- how come the answer isn't simply 4P4?

4⁴

Each person has a choice of 4 rooms, so if there are 4 people, the no. ways they can choose rooms is 4 x 4 x 4 x 4 = 256

33. In how many ways can 5 men and 5 women be arranged in a circle so that the men are separated? In how many ways can this be done if two particular women must not be next to a particular man? (answer = 2880; 864)

First part:
Seat 1 man anywhere (1), then seat the rest (4!), then seat the women (5!)

1 x 4! x 5! = 2880

Second part:

Seat the men (4!), then seat one of the women (she has a choice of 3 spots), then seat the other woman (2 spots to choose from), then seat the rest of the women (3!)

4! x 3 x 2 x 3! = 864

 

[hermand]

33.
either a woman or a man must choose a seat to sit down in. since it is a circle, it does not matter what seat he/she sits in, as they are all the same. there is no beginning or end. however, once she sits down, she defines a beginning. [let's just say for arguments sake she is a woman.]
the woman sits down, leaving four other spaces to be occupied by four other women. therefore 4P4.
there then are five seats in which the five men may sit, so 5P5.
therefore answer is 5P5 x 4P4 = 2880.
----
second part;
have the man sit down first. he does not count as stated before.
this leaves three positions for the two women to sit in, so 3P2. and there are three women and three seats left, so 3P3. there are four positions left for the men to sit in, so 4P4.
3P2 x 4P4 x 3P3 = 864.

 

[scardizzle]

alternativley for 23 i)
e precedes i half of the time
therefore the answer is 7!/2
=2520

 

[hermand]

alternativley for 23 i)
e precedes i half of the time
therefore the answer is 7!/2
=2520

good solution.

i tried to explain it in the most detailed way possible though. so it can be applied to other questions.

 

[Michaelmoo]

23.i.
if e is in the first position, i can be in six positions, and the rest is 5P5.
if e is in the second position, i can be in five positions and the rest is 5P5.
if e is in the third position, i can be in four positions and the rest is 5P5.
etc, so the equation comes out as..

Just a sugestion. There are 7! different arrangements. In all of those, either "e" is before "i" or "i" is before "e". If you consider all the arrangements where "e" is before "i", there will be EXACTLY the same number of arrangments where "i" is before "e" (as you simply interchange e with i)

So "e" precedes "i" in half of the total no. of arrangements. = 0.5 x 7! = 2520

Your reasoning is more mathematically correct though. Dno if they'd accept this.

 

[Smilebuffalo]

28. A car can hold 3 people in the front seat and 4 in the back seat. In how many ways can 7 people be seated in the car if 2 particular people must sit in the back seat and 1 particular person is the driver?

ways in which those 2 people can sit: 4P2 (selecting 2 seats out of 4)
ways in which the other people can sit: 4!
total: 4P2 x 4! = 288

Ahh okay. So in this question do we just disregard the driver since they are confined to one position?

 

[Timothy.Siu]

Just a sugestion. There are 7! different arrangements. In all of those, either "e" is before "i" or "i" is before "e". If you consider all the arrangements where "e" is before "i", there will be EXACTLY the same number of arrangments where "i" is before "e" (as you simply interchange e with i)

So "e" precedes "i" in half of the total no. of arrangements. = 0.5 x 7! = 2520

Your reasoning is more mathematically correct though. Dno if they'd accept this.

thats the way i would do it.

 

[lychnobity]

Ahh okay. So in this question do we just disregard the driver since they are confined to one position?

yes

 

[Smilebuffalo]

23.ii.
working it out similarly to the above question,
if e is in the first position, i can be in three positions and the rest are 5P5.
if e is in the second position, i can be in two positions and the rest are 5P5.
if e is in the third position, i can be in one position and the rest are 5P5.
which makes..

This question seems a bit ambiguous. Did it mean that there has to be exactly 3 letters between e and i? Or did it mean a minimum of 3 letters between e and i? You're solution assumes 3 or more letters apart right? :S

 

[ninetypercent]

Ahh okay. So in this question do we just disregard the driver since they are confined to one position?

yeah you can, but the theory behind it is that you multiply by one

This question seems a bit ambiguous. Did it mean that there has to be exactly 3 letters between e and i? Or did it mean a minimum of 3 letters between e and i? You're solution assumes 3 or more letters apart right? :S

ii) there are 3 letters between e and i (answer = 720)

its exactly three letters. hermand's solution is assuming ONLY 3 letters apart.

 

[lychnobity]

This question seems a bit ambiguous. Did it mean that there has to be exactly 3 letters between e and i? Or did it mean a minimum of 3 letters between e and i? You're solution assumes 3 or more letters apart right? :S

hmm I read the question. Seems it requires exactly 3 letters in between

23. Find the number of arrangements of the letters in the word pencils if:

ii) there are 3 letters between e and i (answer = 720)

There are 3 ways for the e & i to be separated (look down, I think from what you didn't get from hermand's solution is the arrangement can be taken from both ends), 5! to arrange the other letters, and 2 for the e & i to swap places

3 x 2 x 5! = 720

E __ __ __ I __ __
__ E __ __ __ I __
__ __ E __ __ __ I

 

[Smilebuffalo]

nevermind. lychnobity clarified it.

 

[hermand]

yeah you can, but the theory behind it is that you multiply by one



ii) there are 3 letters between e and i (answer = 720)

its exactly three letters. hermand's solution is assuming ONLY 3 letters apart.

no, my sol'n considers anything more than three letters apart. i don't understand why i got the same answer? cause it didn't say exactly three letters apart i assumed it meant at least.

EDIT; realised what i did. cause i assumed e still had to precede i because when i did it that way (was going to do it as their positions being interchangeable too) i got the given answer.

damn being given the answer. it confussed me.