# Discrete Math 101: Fun with Combinatorics – 2

April 17, 2011 Leave a comment

3.

Our task is to select a team of k players from among n players. Now, we have 2 options – Either select player 1 to be part of the team, or leave him out of the team.

If we select player 1 to be part of the team, our task is reduced to selecting players from among players.

And if player 1 is not part of the team, our task is reduced to selecting players from among players.

Hence we have the above result.

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4.

Again our task is to select a team of k players from among n players – .

If is part of the team: We can select such a team in ways.

If is not part of the team, and is part of the team: The problem is reduced to selecting k-1 players from among n-2 players.

If and are not in the team, and is in the team: The problem is reduced to that of selecting k-1 players from among n-3 players.

Proceeding in this manner, we get the above result.

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5.

Again, we have n players at our disposal from among whom we wish to select a team of k players.

Say we select to be the 1st player in our team. (Note that the term “first” actually means nothing since we wish to select a “set” of objects and not an “ordered list”. However the usage of such terminology is useful for the purposes of exposition.)

Now, all we need to do is to select players from among players.

Now, let’s go back a step. What if we select to be the 1st player of the team, and select the remaining players for the team in ways.

We can see that we have n choices for the 1st player of a team.

Hence, we can select a team of k players in .

But again some teams are being repeated.

Given a fixed team, say T, we select that team k times (with each of the k players in the team being selected as the 1st player of the team). Hence we need to divide the previous result by k.

Hence we get the above result.

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