Please check the solutions

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tanmoy
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Please check the solutions

Unread post by tanmoy » Thu Oct 23, 2014 10:25 pm

This is a problem from the book "Combinatorics:A Problem Oriented Approach",by Daniel A. Marcus.
I have solved the problem.But I am little confused about my solution.So,I am posting my solution.Please,check my solution and inform me if I am correct or wrong.

$Problem$:Find the number of ways to place $m$ flags on $n$ distinct poles with at least one flag on each pole if:
(a)the flags are identical.
(b)the flags are distinct.

$Solution$:(a)First place $1$ flag on each pole.$\because$ the flags are identical,it can be done in one way.Now,there are $m-n$ flags remained.The first flag can be placed in any of the $n$ poles.$\therefore$ the first flag can be placed in $n$ ways.The second also can be placed in $n$ ways.Similarly,the $3rd$,the $4th$........the $m-nth$ flags are also can be placed in $n$ ways.So,total $n\times n\times n\times .......\times n_{m-n times}=n^{m-n}$
(b)First place $1$ flag on each pole.$\because$ the flags and the poles both are distinct,it can be done in $P(m,n)$ ways.Now,there are $m-n$ flags remained.The first flag can be placed in $2n$ ways($\because$,there are two places in each flag,either above or below of a flag).Thus,the $2nd$ can be placed in $2n+1$ ways,the $3rd$ can be placed in $2n+2$ ways.......,the $m-nth$ can be in $2n+m-n-1=n+m-1$ ways.So,total $P(m,n)\times 2n\times (2n+1)\times (2n+2)\times ..........\times (n+m-1)$
"Questions we can't answer are far better than answers we can't question"

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