couple questions, perms/comb der (1 Viewer)

currysauce

Actuary in the making
Joined
Aug 31, 2004
Messages
576
Location
Sydney
Gender
Male
HSC
2005
I thought this first one would be easy

8 pink beads and 8 white beads make up a necklace.

how many ways can the necklace be made if;
i) there are no restrictions (2 sig. fig.)

i thought 16! -> but thats the wrong solution??

ii) pink and white alternate

i thought 8!.8! x2

meh


2. A country town uses 6 digits for its telephone numbers with the first 2 digits the same for all the telephone numbers. How many different numbers are possible.
 

acmilan

I'll stab ya
Joined
May 24, 2004
Messages
3,989
Location
Jumanji
Gender
Male
HSC
N/A
For the first one, I think you assume all the pink ones are the same and all the white ones are the same, hence with no restrictions it'd be 16!/8!8!. Same for part 2 of that question
 

KFunk

Psychic refugee
Joined
Sep 19, 2004
Messages
3,323
Location
Sydney
Gender
Male
HSC
2005
A necklace is ussually circular so they're probably ditching the permutations which are identical via rotation.

Is the answer anything like 15!/(8!7!) = 6,435 ?
 

currysauce

Actuary in the making
Joined
Aug 31, 2004
Messages
576
Location
Sydney
Gender
Male
HSC
2005
yep, makes sense too :) necklace is a circle HAHA

- any luck on qu 2.
 

insert-username

Wandering the Lacuna
Joined
Jun 6, 2005
Messages
1,226
Location
NSW
Gender
Male
HSC
2006
I think there's another clause with necklaces, bracelets, and the like: they can be flipped over as well. When you flip it, you halve the number of possible combinations. At least, I think that's right. Also, if there are 8 pink and 8 white, I think there'll be 8! x 8! combinations that look the same. So:


i) there are no restrictions (2 sig. fig.)

(15!/2) ÷ (8! x 8!) = 402.1875 (take to 2 significant figures)


ii) pink and white alternate

You can start with a pink one or you can start with a white one, and they all look the same, so I think 2.


I'm not very sure about that - I'm not too good at perms & combs. Anyone else?


I_F
 

word.

Member
Joined
Sep 20, 2005
Messages
174
Gender
Male
HSC
2005
I think there's another clause with necklaces, bracelets, and the like: they can be flipped over as well.

That's right. Arranging 3 people in a circle = 2! whereas there is only 1 way of arranging 3 beads on a bracelet: 2!/2.

Q2. The question is abit obscure:
If the two numbers are the fixed: 104
Or the first two digits are the same (i.e. 11, 22, 33...): 105
 

mattchan

Member
Joined
Jul 23, 2004
Messages
166
Gender
Male
HSC
2005
currysauce said:
I thought this first one would be easy

8 pink beads and 8 white beads make up a necklace.

how many ways can the necklace be made if;
i) there are no restrictions (2 sig. fig.)

i thought 16! -> but thats the wrong solution??

ii) pink and white alternate

i thought 8!.8! x2

meh


2. A country town uses 6 digits for its telephone numbers with the first 2 digits the same for all the telephone numbers. How many different numbers are possible.

Lets see...

i) 15! /(8! * 8!)
ii) 7! *8! / (8! * 8!)
iii) 10 * 10 * 10 * 10 = 10^4

Probably wrong but meh
 

Pace_T

Active Member
Joined
Oct 21, 2004
Messages
1,784
Gender
Male
HSC
2005
i) 15!/(7!8!)
ii) 7!8!/(7!8!) = 1
think about it, how many ways can u actually arrange 16 of 2 different coloured beads in a circle while they are alternating? lol.. i think.
2) [ 1*10 ] * 10 * 10 * 10 * 10 = 10^5
 

KFunk

Psychic refugee
Joined
Sep 19, 2004
Messages
3,323
Location
Sydney
Gender
Male
HSC
2005
Pace_T said:
ii) 7!8!/(7!8!) = 1
think about it, how many ways can u actually arrange 16 of 2 different coloured beads in a circle while they are alternating? lol.. i think.
That's what I figured. '1' by common sense.
 

currysauce

Actuary in the making
Joined
Aug 31, 2004
Messages
576
Location
Sydney
Gender
Male
HSC
2005
and these questions come from Maths in Focus...

anyway

the answers were

1)

i)6.5 x 10 ^11
ii) 101 606 400

2.

10000
 

insert-username

Wandering the Lacuna
Joined
Jun 6, 2005
Messages
1,226
Location
NSW
Gender
Male
HSC
2006
ii) 101 606 400

That's impossible. Unless you do something like number the beads so they look different. Ack.

I've got the Maths in Focus textbooks. Where's the question - I'll have a look in mine and see if it has the same.


I_F
 
Last edited:

word.

Member
Joined
Sep 20, 2005
Messages
174
Gender
Male
HSC
2005
mattchan said:
With 1) i). 15!/2. Why do you divide by 2?
Because a necklace is a 3D object which can be flipped with the 'fixed' bead in the same position hence halving the possible number of arrangements.
Graphically:
 
Last edited:

woho

Member
Joined
Sep 9, 2006
Messages
49
Gender
Female
HSC
2007
the question is Revision set 4 23b)

Amazingly, [(8!*8!) ÷ 16] gives the answer
 
Last edited:

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

Top