Probability and Combinations

Question

show that

1)5 ( n ) = n ( n - 1 )

5 4

2) r ( n ) = n ( n - 1 )

r r - 1

3) 2 ( 2n -1 ) = ( 2n )

n-1 n

*the bracket includes the top and the bottom row

Answer 1

I'll work through problem 1 - you can use almost identical reasoning for the remainder of the problems.

Approach 1: Algebraic.

Expanding the left-hand side, we get

5n!/(5! (n-5)!) = n!/ (4! (n-5)!) (canceling a factor of 5 in the numerator and denominator).

= n(n-1)! / (4! (n-5)!) = n (n-1 choose 4).

Approach 2: Reasoning.

Choosing 5 objects from a selection of n objects is equivalent to choosing 1 object from a selection of n objects, and then choosing 4 objects from the remaining n - 1 objects. However, n(n-1 choose 4) overcounts by a factor of 5, since we don't care about which object gets chosen first. Therefore, 5(n choose 5) = n (n-1 choose 4).