The number of ways we can make team if Steve and Danny insist on playing on opposing teams will be 70.
We have,
Total number of team = 01,
And,
Total number of players = 10,
And,
Have to divide team in group of two of 5 player each,
So,
Now,
According to the question,
Number of players left after having Denny and Steve on opposing sides = 10 - 2 = 8
Now,
As Denny and Steve on opposing sides,
So,
i.e. n = 8,
And
r = 4
The number of ways team can be arranged will be ⁸C₄, i.e. using Combination formula,
i.e.
[tex]{^n}C_{r}=\frac{n!}{r1(n-r)!}[/tex]
i.e.
Putting values,
⁸C₄ = [tex]\frac{8!}{4!(8-4)!}[/tex]
On solving we get,
⁸C₄ = [tex]\frac{8*7*6*5*4!}{4*3*2*1*4!}[/tex]
On solving further we get,
⁸C₄ = 2 × 7 × 5
i.e.
⁸C₄ = 70
So, teams can be arranged in 70 ways.
Hence we can say that the number of ways we can make team if Steve and Danny insist on playing on opposing teams will be 70.
Learn more about number of ways herehttps://brainly.com/question/13003667
#SPJ4
