((n choose 2) choose 2) = 3 . (n choose 4) + 3 . (n choose 3)
prove with a combinatorial argument.
why people hate prove algebraically.
thanx
Dear friend,
Let us consider a game cricket in which each team has n players.(Generally there are 11 players but for our problem we define a new cricket game.)
Indian government decides to give two awards for two pairs of opening batsmen. So we are trying to find no. of possibilities to give this award. First, the no. of pairs are: (n choose 2). Now we have to choose 2 of these for award. Hence there are {(n choose 2) choose 2} chances.
Now let us look at another way to solve this problem.
There are 2 possibilities: 1) The 2 pairs selected for awards consist 4 different players. Let the players be A, B, C, D. Now awards can be given to these players in 3 following ways.
A) Pair 1: A, B Pair 2: C, D.
B) Pair 1: A, C Pair 2: B, D.
C) Pair 1: A, D Pair 2: B, C.
Hence there are 3*(n choose 4) such ways.
Second possibility: One player is common to the two selected pairs. i. e. only 3 different players will get the award. Let the players be A, B, C. Now we have to decide that which player should be repeated hence again there are 3 ways.
A) Pair 1: A, B Pair 2: B, C.
B) Pair 1: A, B Pair 2: A, C.
C) Pair 1: A, C Pair 2: B, C.
Hence there are 3*(n choose 3) such ways.
Hence total ways are 3*(n choose 3) + 3*(n choose 4).
The answers by these two methods should be same.
Hence 3*(n choose 3) + 3*(n choose 4) = {(n choose 2) choose 2}.
Please reply me by sending an email to me after you read this solution. My email ID is ajinkya1991@rediffmail.com.It is bet...eply.<br />
-Ajinkya Patil.