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November 26, 2014

November 26, 2014

Posted by **Alex** on Sunday, January 29, 2012 at 5:27pm.

- Math -
**Reiny**, Sunday, January 29, 2012 at 6:20pmVisualize the beads on a string stretched out in front of you

Cases:

A) all blue --- one way

B) 1 red, 4 blue:

number of choices = 5!/(1!4!) = 5

e.g.

RBBBB, BRBBB, BBRBB, BBBRB, BBBBR

BUT... RBBBB and BBBBR are really the same bracelet, merely flipped, so there are really only 3 of these

C) 2 red, 3 blue

number of choices = 5!/(2!3!) = 10

again , 5 of those are symmetrical, so only 5 of these

D) 3red, 2 blue

same as C), --- so 5 of those

E) 4red, 1 blue

same as B), so 3 of these

F) all red ---- one of them

total = 1 + 3 + 5 + 5 + 3 + 1 = 18

check my thinking.

- Math -
**Christine**, Monday, January 30, 2012 at 1:20am8. If all 5 beads are red, only 1 bracelet can be formed. If 4 beads are

red and 1 is blue, only 1 bracelet can be made. If 3 beads are red and 2 beads are blue, 2 different bracelets are possible: the blue beads together or separated by the red beads. The symmetry in this situation means that there will be 2 bracelets with 2 red and 3 blue beads, 1 bracelet with 1 red bead, and 1 bracelet with no red beads, for a total of 8 bracelets.

- Math -
**jenny**, Wednesday, October 3, 2012 at 10:37pma bracelet is to be made from 2 red and 5 blue beads. how many different bracelets can be made

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