julie plans to take part in the office betting pool. Each participant must choose the first...

julie plans to take part in the office betting pool. Each participant must choose the first second and third place winners in a basketball tournament and write them on their ticket. If julie chooses correctly, she will win $1000. If there are 28 basketball teams to choose from in the tournament, how many different tickets are possible?

Solutions

Expert Solution

We can use concept of permutation as we can see

While selecting first place winner we have 28 teams

After selecting first place winners we have to choose from 27 remaining teams our second place winners

After selecting second place winners we remain with 26 teams for selecting third place winners

So the total possibilities can be = 28*27*26 = 19656

We can also use permutation formula which is

P(n,r) = n!/(n-r)! where in our question n=28 and r=3 (n! = n*(n-1)*(n-2).....)

=28!/(28-3)! = 28!/25! =19656

In both the ways answer will be same

So total of 19656 tickets can be possible.

Rahul Sunny answered 2 years ago

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