couple questions, perms/comb der (2 Viewers)

[currysauce]

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]

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

 

[currysauce]

nope, not what the book says

 

[KFunk]

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 ?

 

[word.]

Um,

i) 15!/2
ii) 8!7!/2

I think...

 

[currysauce]

yep, makes sense too necklace is a circle HAHA

- any luck on qu 2.

 

[insert-username]

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.]

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]

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

 

[insert-username]

ii) 7! *8! / (8! * 8!)

1/8 of a way?


I_F

 

[mattchan]

Nevermind my foolishness =)

 

[Pace_T]

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]

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]

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]

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

 

[mattchan]

With 1) i). 15!/2. Why do you divide by 2?

 

[word.]

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:

 

[woho]

the question is Revision set 4 23b)

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