## Problem Of The Week 9

**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?**

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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

milindSeptember 11, 2008 at 4:40 pm

You misunderstood the problem, please try again at milind !

outtimedSeptember 11, 2008 at 4:50 pm

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.

TanujSeptember 12, 2008 at 5:38 am

you have messed it up @ tanuj try again !

outtimedSeptember 12, 2008 at 6:08 am

The problem is basically to arrange aaabbbcc

and which can be done in an 8!/3!3!3!=560 ways

outtimedSeptember 12, 2008 at 5:55 pm