Your making a necklace with 4 red peices and 4 yellow peices. How many different combination of necklaces can you make?

is 2^8 correct.

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- Oct 29th 2008, 03:41 PMharold3331combinations
Your making a necklace with 4 red peices and 4 yellow peices. How many different combination of necklaces can you make?

is 2^8 correct. - Oct 30th 2008, 12:53 PMBoboStrategyRe: necklace
2^8 is not correct.

Here's why:

2^8 assumes a necklace like:

RYRYRYRY

is different than a necklace like:

YRYRYRYR

You need to reflect this in your counting to obtain the correct (i.e. lower) answer. - Oct 30th 2008, 03:50 PMSoroban
Hello, harold3331!

This is a**very**tricky problem . . .

Quote:

You're making a necklace with 4 red pieces and 4 yellow pieces.

How many different combination of necklaces can you make?

is 2^8 correct? . . . . no

Your answer assumes there are at least eight of each color

. . and includes all the combinations from $\displaystyle RRRRRRRR\text{ to }YYYYYYYY.$

Even worse, this is a necklace . . . a__circular__arrangement.

So that, for example: .$\displaystyle RRRRYYYY \:=\:RRYYYYRR$

. . That is: . $\displaystyle \begin{array}{cccc}& R\;R \\ Y & & R \\ Y & & R \\ & Y\;Y \end{array}\quad=$ . .$\displaystyle \begin{array}{ccc}& R\;R \\ R & & Y \\ R & & Y \\ & Y\;Y \end{array}$

There is one more "trap" in a Necklace Problem, overlooked by the best of us.

This is similar to seating 4 men and 4 women around a circilar table,

. . except a necklace can be "flipped over".

So that: . $\displaystyle \begin{array}{ccc} & R:R \\ Y &: & R \\ Y &:& Y \\ & Y:R\end{array}$ . .$\displaystyle = \quad \begin{array}{ccc}& R:R \\ R &:& Y \\ Y &:& Y \\ & R:Y\end{array}$

Care to try again?