Math

Red and blue beads are being strung to make a bracelet. How many different bracelets can be formed it each bracelet uses exactly 5 beads?

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  1. Visualize 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.

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  2. 8. 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.

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  3. a bracelet is to be made from 2 red and 5 blue beads. how many different bracelets can be made

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  4. three

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  5. Blue=25 Red=30

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