Best Answer

The general formula for the number of combinations of r things chosen from n different things is C(n,r) = n!/ r!(n-r)! How many ways can 5 players be chosen from a team of 13?

(Where the order you choose the players does not matter).

So C(13,5) = 13!/5!(13-5)!

Or 1,287

More answers

There are 120 possible ways.

To calculate the number of permutations here, where order is important and repetition is not allowed, we use the following formula:

Number of permutations = n! / (n - r)! = 5!/0! = 1*2*3*4*5 / 1 (note: "!" means "factorial" and 0! equals 1) = 120/1 = 120.

For a complete list see below. Let a,b,c,d & e represent the 5 players and their order determine their position:

{a,b,c,d,e} {a,b,c,e,d} {a,b,d,c,e} {a,b,d,e,c} {a,b,e,c,d} {a,b,e,d,c} {a,c,b,d,e} {a,c,b,e,d} {a,c,d,b,e} {a,c,d,e,b} {a,c,e,b,d} {a,c,e,d,b} {a,d,b,c,e} {a,d,b,e,c} {a,d,c,b,e} {a,d,c,e,b} {a,d,e,b,c} {a,d,e,c,b} {a,e,b,c,d} {a,e,b,d,c} {a,e,c,b,d} {a,e,c,d,b} {a,e,d,b,c} {a,e,d,c,b} {b,a,c,d,e} {b,a,c,e,d} {b,a,d,c,e} {b,a,d,e,c} {b,a,e,c,d} {b,a,e,d,c} {b,c,a,d,e} {b,c,a,e,d} {b,c,d,a,e} {b,c,d,e,a} {b,c,e,a,d} {b,c,e,d,a} {b,d,a,c,e} {b,d,a,e,c} {b,d,c,a,e} {b,d,c,e,a} {b,d,e,a,c} {b,d,e,c,a} {b,e,a,c,d} {b,e,a,d,c} {b,e,c,a,d} {b,e,c,d,a} {b,e,d,a,c} {b,e,d,c,a} {c,a,b,d,e} {c,a,b,e,d} {c,a,d,b,e} {c,a,d,e,b} {c,a,e,b,d} {c,a,e,d,b} {c,b,a,d,e} {c,b,a,e,d} {c,b,d,a,e} {c,b,d,e,a} {c,b,e,a,d} {c,b,e,d,a} {c,d,a,b,e} {c,d,a,e,b} {c,d,b,a,e} {c,d,b,e,a} {c,d,e,a,b} {c,d,e,b,a} {c,e,a,b,d} {c,e,a,d,b} {c,e,b,a,d} {c,e,b,d,a} {c,e,d,a,b} {c,e,d,b,a} {d,a,b,c,e} {d,a,b,e,c} {d,a,c,b,e} {d,a,c,e,b} {d,a,e,b,c} {d,a,e,c,b} {d,b,a,c,e} {d,b,a,e,c} {d,b,c,a,e} {d,b,c,e,a} {d,b,e,a,c} {d,b,e,c,a} {d,c,a,b,e} {d,c,a,e,b} {d,c,b,a,e} {d,c,b,e,a} {d,c,e,a,b} {d,c,e,b,a} {d,e,a,b,c} {d,e,a,c,b} {d,e,b,a,c} {d,e,b,c,a} {d,e,c,a,b} {d,e,c,b,a} {e,a,b,c,d} {e,a,b,d,c} {e,a,c,b,d} {e,a,c,d,b} {e,a,d,b,c} {e,a,d,c,b} {e,b,a,c,d} {e,b,a,d,c} {e,b,c,a,d} {e,b,c,d,a} {e,b,d,a,c} {e,b,d,c,a} {e,c,a,b,d} {e,c,a,d,b} {e,c,b,a,d} {e,c,b,d,a} {e,c,d,a,b} {e,c,d,b,a} {e,d,a,b,c} {e,d,a,c,b} {e,d,b,a,c} {e,d,b,c,a} {e,d,c,a,b} {e,d,c,b,a}

Math I know; Basketball I'm a little weaker on.

Assuming that each of the five positions is distinct there are 8! / 3! or 6720 possible combinations.

Assuming that all of the five positions are the same (I know this isn't true; I'm just putting it up for comparison) there are 8 choose 5 ( 8!/(3!*5!)) or 56 possible combinations.

If some of the positions are the same but others aren't, then the answer is somewhere in between.

Lvl 2

Some physical properties of five different substances (A, B, C, D and E) are outlined in the table below. Which of the substances are nonmetals? *

twenty one

Q: How many ways can the 5 starting positions on a basketball team be filled 8 men who can play any of the positions?

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