Permutations (1 Viewer)

[DcM]

Hey

i dunno how to do this question from 3u Fitzpatrick, its a permutation question..

Q.29 In how many ways can 4 people be accommodated if there are 4 rooms available?


i dun understand permutations at all

thanks

 

[wogboy]

There's something missing here, can each room accommodate:

a) an unlimited amount of people?
b) or is it only one person to a room?


case a)
Each person can choose from 4 rooms. There are 4 people. Think of it like this, each person is like a digit placeholder, and each room is a digit (from 1-4). How many numbers can you form? (note that there may be more than 1 person in a particular room)

4^4 = 256 ways they can be arranged.

case b)
4 people can choose from 4 rooms, and the rooms can't be shared. The order here obviously DOES matter (John-Room1 & James-Room2 is different from James-Room1 & John-Room2) . So there are:

4P4 = 4! = 24 ways.

 

[KeypadSDM]

Originally posted by wogboy
The order here obviously DOES matter

I dunno dude, the rooms may not be numbered, they may all be in a circle.

 

[Grey Council]

im quite sure its case a . . . i remember (vaguely) doing this.

Stuff that
*takes out fitzpatrick book*
question 29, exercise 28(a)
answer: 256

I was right.

BTW, one can usually work out which case it is. In this question, it was question 29, I really doubt theyd give a simple 4P4 question as a question 29. And don't worry, in the exams you won't have ambiguosly worded questions. So as long as you understand how to do it, imho, it doesn't really matter if you can't do one question in the middle of an exercise.

 

[DcM]

Hey thanks for helping me out

got into another question..from same book ex 28c q.6

Five cards are drawn from a pack of 52 playing cards. What is the probability of drawing at least 3 aces?

how wud u approach these kinda of question?

 

[clerisy]

Here's how I would do it, but PLEASE tell me if I'm wrong. I've kind of neglected permutations/combinations and its one of my worse topics...


Total number of choices is 52C5 (combination, not permutation because order doesn't matter) = 2,598,960

Total number of choices with at least 3 aces is (I think) 49C2-- because after the three aces are taken from 52 cards there are obviously 49 left, and if youre drawing 5 cards there are only 2 spaces left.
= 1,176

Therefore probability is 1,176 / 2,598,960

= 1 / 2210

 

[DcM]

Um...well from the book the answer is..
19/10829

so yea?? help? ><

 

[DcM]

i dunno if your way to do it is right clerisy
cos does 49C2 just mean the remaining cards..so wat about the aces..
and also at least 3 aces..so i think that means u can have 4 aces too....

but then i keep on trying to do it...but i dun get it right! -_-'
Help.....

 

[Heinz]

(4C3 x 48C2 + 4C4 x 48C1) / 52C5 = 19/10829

 

[clerisy]

I see.

hmm. I think that I will go over some permutations stuff before I go back to school!

 

[Grey Council]

heaps of perms and combs questions (esp probability) you hafta do in parts. Remeber that, if your way doesn't work, then do individual cases. As Heinz did (if you didn't understand it, although i'm sure you did if you got up to question 29 without any problems )
3 of the aces x any other 2 remaining cards + 4 of the aces x any other card. I dunno how else to explain it.

 

[victorling]

Originally posted by DcM
Hey

i dunno how to do this question from 3u Fitzpatrick, its a permutation question..

Q.29 In how many ways can 4 people be accommodated if there are 4 rooms available?


i dun understand permutations at all

thanks

dont worry
fitzpatrick is the best book for perm and comb
just practice and do the questions over and over again

 

[victorling]

by the way
the answer is 4^4

 

[victorling]

because 4 people can be arranged 4 different ways(rooms)

 

[+:: $i[Q]u3 ::+]

fitzpatrick's pretty good - if u've finished those exercises, flick through the test papers in those books - u can find a fair number of perm/com questions that really help u get going.

i hated this topic.. had to do like three freakin million questions until i could get them right...