Question

formula for determining the number of bye in the lower half of a knockout fixture when the number of byes are odd?

​

Answer 1

Answer:

Explanation:

It is very easy. Suppose you have to draw a fixture of 11 teams on knock put basis.

Here we go with the formula:

First divide the teams into two halves, i.e upper half & lower half, then:

No. Of teams in the upper half= N+1/2,

i.e 11+1/2 =12/2 =6 teams in upper half

Then lower half= N-1/2,

= 11–1/2 = 10/2=5 teams in lower half

Therefore total No. Of byes:

Total no. of teams = 11

Next highest power of two =( 2)^4 =16

(Formula= 2^n - N) where N is the no. Of teams.

So the no. Of byes to be placed is

16–11 = 5 byes.

Respond if you have Understood.

Thankyou

Answer 2

It's very easy. Let's say you have to draw an 11 team draw based on knocks.

Here is the formula:

First, divide the teams into two halves, i.e. top half and bottom half, then:

Number of teams in the top half = N+1/2,

ie 11+1/2 =12/2 =6 teams in the top half

The bottom half = N-1/2,

= $(11-1)/2$ = 10/2 = 5 teams in the bottom half

Therefore, the total number of byes:

Total number of teams = 11

Next highest power of two =$2^4$ =16

(Formula= $2^n - N$) where N is a number. From the teams.

So no. Byes are placed in them

16–11 = 5 byes.

As a general rule, in an elimination tournament, the number of byes is calculated by subtracting the total number of teams from the next higher number to the power of 2.

When the number of teams is odd, then the number of byes in the lower half is calculated according to the formula Nb +1/2

brainly.in/question/47708342

#SPJ2