Answer
a) 210
b) 1260
c) 252
Explanation
The problem of choosing a number of options from a bigger number of options with the order of the choices not important is solved using Combination.
The number of ways for picking r options from n choices with order not important is given as
ⁿCᵣ
[tex]^{n}Cᵣ=\frac{n!}{(n-r)!r!}[/tex]a) n = 10 players
r = 4 players
[tex]\begin{gathered} ^{10}C_4=\frac{10!}{(10-4)!4!} \\ =\frac{10\times9\times8\times7\times6!}{6!\times4!} \\ =\frac{10\times9\times8\times7}{4\times3\times2} \\ =210\text{ ways} \end{gathered}[/tex]b) n₁ = 10, n₂ = 8 (after the first two have been selected, there are 8 options left)
r₁ = 2; r₂ = 2
[tex]\begin{gathered} \\ =^{10}C_2\times^8C_2 \\ =\frac{10!}{(10-2)!2!}\times\frac{8!}{(8-2)!2!} \\ =\frac{10!}{8!2!}\times\frac{8!}{6!2!} \\ =\frac{10\times9\times8!}{8!\times2}\times\frac{8\times7\times6!}{6!\times2} \\ =\frac{10\times9}{2}\times\frac{8\times7}{2} \\ =45\times28 \\ =1260 \end{gathered}[/tex]c) n₁ = 10, n₂ = 5 (after the first two have been selected, there are 5 options left)
r₁ = 5, r₂ = 5
So, the number of ways will be equal to
= ¹⁰C₅ × ⁵C₅
= [10!/(10 - 5)!5!] × [5!/(5 - 5)!5!]
= [10!/5!5!] × [5!/0!5!]
Noting that 0! = 1
[10!/5!5!] × [5!/0!5!]
= 252 × 1
= 252 ways
Hope this Helps!!!
