You are playing a game on T20 World Cup. There are $12$ teams and each team has $5$ Batsmen, $5$ Bowlers, $4$ All-rounders and $1$ wicket keeper in their squad.
You have to create a team of $12$ players; consists of $4$ Batsmen, $3$ Bowlers, $3$ A.R , $1$ W.K and a $12^{th}$ man.[The 12^{th} man can be any of the remaining player]. The condition is- you can select only one player from each team.
How many different teams can be created ?
I made this problem. But I'm not sure about the answer.
This problem can be made tougher changing the condition. But it already seems TOUGH! !
Fantasy Cricket
- Fahim Shahriar
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Name: Fahim Shahriar Shakkhor
Notre Dame College
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- nafistiham
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Re: Fantasy Cricket
I think you have to define a serial of choosing them.
Suppose, if you choose in the serial w.k.,a.r.,batsmen and bowler
then, it will be different from the order a.r.,batsmen,bowler and w.k.
Suppose, if you choose in the serial w.k.,a.r.,batsmen and bowler
then, it will be different from the order a.r.,batsmen,bowler and w.k.
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
- Fahim Shahriar
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Re: Fantasy Cricket
I should mention it earlier.
I chose this serial. --Batsmen,Bowler,A.R , W.K and at last $12^{th}$ man--
I chose this serial. --Batsmen,Bowler,A.R , W.K and at last $12^{th}$ man--
Name: Fahim Shahriar Shakkhor
Notre Dame College
Notre Dame College
- nafistiham
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Re: Fantasy Cricket
OK. thanks for that.
solution:
multiply them
by the way $ _{m}^{n}\textrm{P}=\frac {n!}{(n-m)!}$
solution:
by the way $ _{m}^{n}\textrm{P}=\frac {n!}{(n-m)!}$
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.