Permutations with equal amount of e and o

Question

I have a follow up. How would I calculate the e and o combinations that have an equal amount of e’s and o’s in any combination for example with 2 to the forth power there are 16 combinations but only eeoo ooee eoeo oeoe oeeo eooe. How would I find this out for bigger sets.

Answer 1

 

N = 2^k where k is the number of pairs of (0,e)

D = N/2

 

The number of permutations is then  n! /  [(d!)(d!)]

 

This is the same formula as the permutations where there are 2 or more items that are the same.

 

For example, the first pair, k=1, the ordered pair is (o,e)

2^k = 2^1 = 2

D=2/2 = 1

2! / (1!)(1!) = 2

So there are two permutations: oe and eo

 

 

For k=2,  2^2 = 4.

 D=4/2 = 2;

So there are 4!/[(2!)(2!)] = (4*3)/(2*1) = 6 permutations of ooee:

 ooee

 oeoe

 eooe

 eeoo

 eoeo

 oeeo

 

So for k=4 pairs of (o,e) namely the permutations of ooooeeee

there are

 2^4 = 16 ---> D=16/2 = 8

 

(16!)/[(8!)(8!)] = 

 

 (16*15*14*13*12*11*10*9)/(8*7*6*5*4*3*2) =  <---- 8! cancels

 

  (15*14*13*12*11*10*9)/(7*6*5*4*3) =  <--- 16 cancels 8x2

  

  (14*13*12*11*10*9)/(7*6*4)  <----- 15 cancels 5*3

 

  (2*13*2*11*5*9) / 2 <--- 14 cancels 7,  12 cancels 6, 10 cancels 4 by a factor of 2

 

  13*2*11*5*9  <---- one pair of 2's cancel out

 

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