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Problem Of The Week 9

with 5 comments


In a shooting match, eight clay targets are arranged in two hanging columns of three targets each and one column of two targets. A marksman is to break all the targets according to the following rules:

1) The marksman first chooses a column from which a target is to be broken.
2) The marksman must then break the lowest remaining target in the chosen column.

If the rules are followed, in how many different orders can the eight targets be broken?

Written by Implex

September 11, 2008 at 11:46 am

5 Responses

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  1. we can consider this to be an arrangement as follows:

    o o o
    o o o
    o o
    Now, we have to select the bottom most from any column. Sso, the number of distinct arrangements of shooting all will be 3*3*2=18

    milind

    September 11, 2008 at 4:40 pm

  2. You misunderstood the problem, please try again at milind !

    outtimed

    September 11, 2008 at 4:50 pm

  3. Selecting a column out of the three columns (say the one with 3 targets) and then shooting the targets of the column in 3! ways considering them nonidentical as nothing is mentioned.

    3C1*3!.
    out of the remaining two selcting another column in 2C1 ways and then shooting targets in 3! ways and then shooting the remaining targets in 2 ways.

    3C1*3! * 2C1*3! * 2! = N(say)
    Now, if we would have selected the other two in the very first selection then :
    N*3! = P, but
    2 of the columns have same number of balls.
    P/2!
    = 1296.

    Identical targets
    But if the columns had identical target =>
    then no of ways of selecting coloumn =
    3C1 * 1 (since 1 way of shooting three identical target)
    * 2C1 * 1 * 3!/2! = 18.

    Tanuj

    September 12, 2008 at 5:38 am

  4. you have messed it up @ tanuj try again !

    outtimed

    September 12, 2008 at 6:08 am

  5. The problem is basically to arrange aaabbbcc
    and which can be done in an 8!/3!3!3!=560 ways

    outtimed

    September 12, 2008 at 5:55 pm


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