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The Number of ways in which a team of eleven players can be selected from 22 players including 2 of them and excluding 4 of them is:

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Prem: Are you just asking one homework question after the other? Also, can you please look at how other people format their posts? Nobody puts them bold and in italics (and I'm growing tired of reformatting and re-tagging each of your questions). –  t.b. Apr 22 '11 at 13:56
@Theo Buehler....Ha Ha Actuly i am solving question based on Permutation and combination...but i felt difficulty in some of the question thats why i that its worthy to discuss with other Brilliant people around the world.....I know the community rules about the standerd of question.......If i am wrong in any way please let me know...i am very happy to be with the stackexchange...i wlike to listen if any concern.... –  prem shekhar Apr 22 '11 at 14:00
Oh..i will keep in mind from next time surely...... –  prem shekhar Apr 22 '11 at 14:01
Hover your mouse over the tags: e.g. the short description of (number-theory) is "Questions on more advanced topics of number theory" (which your question certainly isn't). I think the appropriate tag for your questions is (combinatorics), have a look at the wikipedia page. –  t.b. Apr 22 '11 at 14:06
@prem: read this and follow the suggestion there. I am with Theo Buehler on this: your questions are formulated poorly with zero motivation. Further abuse will result in suspension of your account. –  Willie Wong Apr 22 '11 at 14:10

1 Answer 1

I'd say that this is equivalent to choosing $9 (=11-2)$ players among $18 (=22-4)$

So : $\binom{18}{9}$

Edit : An error has occured, look at the commentaries, it's $\binom{16}{9}$ in fact

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Except that it is not: on this basis (positions do not matter) it should be choosing 9(=11−2) players among 16(=22−4-2) so ${16 \choose 9}$ –  Henry Apr 25 '11 at 17:26
Oops, you're definitely right =) –  Fezvez Apr 28 '11 at 8:11

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