Question

the table shows the year in school and the number of years experience for 18 players on a high school soccer team. the coach can choose 2 players to be captains. a captain must be a junior (jr) or a senior (sr), and have played on the team for more than 1 year.
how many different captain pairs can the coach choose?
the coach can choose
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different pairs as captains.

player	year in school	years on team
2	fr	1
3	jr	2
4	so	2
5	so	2
6	sr	3
7	sr	4
8	jr	3
9	jr	1
10	fr	1
11	sr	1
12	sr	3
13	sr	4
14	so	2
15	jr	3
16	jr	1
17	sr	2
18	so	2

Explanation:

Step1: Identify eligible players

First, count players who are Jr/Sr AND have >1 year on team:

• Player1: Sr, 1yr → no
• Player2: Fr, 1yr → no
• Player3: Jr, 2yr → yes
• Player4: So, 2yr → no
• Player5: So, 2yr → no
• Player6: Sr, 3yr → yes
• Player7: Sr, 4yr → yes
• Player8: Jr, 3yr → yes
• Player9: Jr, 1yr → no
• Player10: Fr, 1yr → no
• Player11: Sr, 1yr → no
• Player12: Sr, 3yr → yes
• Player13: Sr, 4yr → yes
• Player14: So, 2yr → no
• Player15: Jr, 3yr → yes
• Player16: Jr, 1yr → no
• Player17: Sr, 2yr → yes
• Player18: So, 2yr → no

Eligible players: 3,6,7,8,12,13,15,17 → total of 8 players.

Step2: Calculate pair combinations

Use combination formula $\binom{n}{k}=\frac{n!}{k!(n-k)!}$, where $n=8$, $k=2$.
$$\binom{8}{2}=\frac{8!}{2!(8-2)!}=\frac{8\times7}{2\times1}=28$$

Answer:

28