A tennis club has 9 members who play singles. The club must fill three different roles (captain, vice-captain and social secretary) with three different members chosen from the 9. How many different ways can these three roles be filled?

AS Level Maths 9709: Permutations, Filling Distinct Roles (nPr)

Step 1: Decide whether order matters

The three roles (captain, vice-captain, social secretary) are all different. Assigning Priya as captain and Reza as vice-captain is a different outcome from assigning Reza as captain and Priya as vice-captain. So order matters, and this is a permutation, not a combination.

Step 2: Count directly, role by role

Captain: any of the $9$ members can be chosen, $9$ ways.
Vice-captain: one member has already been used up, so $8$ members remain, $8$ ways.
Social secretary: two members have already been used, so $7$ members remain, $7$ ways.

By the multiplication principle: $9\times8\times7 = 504.$

Step 3: Check using the permutation formula

$^9P_3 = \frac{9!}{(9-3)!} = \frac{9!}{6!} = 9\times8\times7 = 504.$

Both methods agree, confirming $504$ .

Why the other options are wrong

A ( $84$ ): this is $\binom{9}{3}$ , the number of ways to select 3 members with no distinct roles. It ignores that the roles are different.
C ( $729$ ): this is $9^3$ , which would be correct only if the same member could hold more than one role.
D ( $362880$ ): this is $9!$ , the number of ways to arrange all 9 members in order, not just fill 3 roles.

Final answer

Number of ways to fill the three roles $= \boxed{504}$ , option B.