How many ways can four numbers be drawn from a group of ten numbers if the order does not matter?

Answer:210 waysStep-by-step explanation:Simply put in combinations and permutations,ORDER MATTERS >> PERMUTATIONORDER DOESNT MATTER >> COMBINATIONFormula for Permutation is  [tex]P(n,r)=\frac{n!}{(n-r)!}[/tex]Formula for Combination is  [tex]C(n,r)=\frac{n!}{(n-r)!r!}[/tex]These means taking r objects from a group of nAlso x! means x(x-1)(x-2)... (ex: 4! = 4 * 3 * 2 * 1)Now, from the question, we have "order doesn't matter" so we have combination with n = 10 and r = 4.Let's put the numbers into formula and find the answer:[tex]C(n,r)=\frac{n!}{(n-r)!r!}\\C(10,4)=\frac{10!}{(10-4)!4!}\\=\frac{10!}{6!4!}\\=\frac{10*9*8*7*6!}{6!4!}\\=\frac{10*9*8*7}{4*3*2*1}\\=210[/tex]So, the answer is 210