The number of teams that can be chosen from 10 players could be rather large, so one reason to answer this question is to decide whether or not it would be a good idea to try to choose the players for one's team by making a complete list of all the possible teams.and then deciding which one looks best.
Here's how we figure it out.
We can consider the number of possible different teams by finding the number of different orderings (or permutations) of our 7 choices. There are 10 players available so there are 10 choices for our first pick. There would be 9 players still available (since we can't just keep picking the same player, he or she is no longer available to be picked) so there are 9 choices for our second pick. Similarly, there would be 8 choices for our 3rd pick, and you see the pattern is that there are 11-n choices for our nth pick so that there are still four players to choose from for our seventh and final pick. Altogether, all of these number of choices multiply to produce all the different choices if every different ordering counts as a different choice, even though many of the picks would include the same 7 players, just in a different order. So the number of different permutations ends up being 10*9*8*7*6*5*4. Notice that this product of descending integers stops at 4 (the 7th number). We could express this result more compactly as 10!/3! where we pronounce 10! as "ten factorial" and "10 factorial" or 10! means 10*9*8*7*6*5*4*3*2*1 and 3! means 3*2*1 . Our quotient of these two factorials is equivalent to the number of permutations expression (10*9*8*7*6*5*4) because the 3*2*1 factors cancel. This formula is often available in calculators under the name "nPr" (Permutations from n items with r choices) and is equivalent to n ! / ( n - r) !
We are on the right track but have overcounted by quite a bit because we were considering every different ordering of the same 7 players as a different possibility. If we think about how many different orderings the same 7 players can be chosen in, we see there are 7 possible players to pick first, 6 to pick second, etc.So the number of orderings that are actually still the same team selection would be 7!.
Therefore the number of different orderings is given by a new symbol nCr which is pronounced "n choose r" and n is the number of items from which to choose and r is the number of items selected, whenever we do not want to include in our count of the possibilities different orderings of the same selection. This formula is equivalent to nPr (or n!/r!) divided by the number of orderings of the selection (r!).
nCr = n !
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( n -r ) ! r !
In our case, then, the number of ways to choose 7 players from 10 available is
10!/ ( (3!)·(7!) = 10 · 9 · 8 7! 5 · 2 · 3 · 3 · 8 120 combinations available from 10 players.
------------------- = ------------------------ =
7! 3 · 2 · 1 3 · 2
Notice this problem can also be viewed as "how many ways are there to put 3 players on healthy reserve?"
since nCr is equivalent to nCn-r!
