Different starting lineups can be created - 120+60+120=300. A certain college team has on its roster 4 centers, 5 guards, 5 forwards, and one individual (x) who can play either guard or forward.
1. A lineup without x is an example. You must decide.
- two of the three forwards;
- from 4 centers, 1 center.
It can be made as -
2C5 × 2C3× 1C4 = 10× 3×4 = 120 ways
2. Consider starting lineups with guard x. You must decide.
- 2 guards chosen from 6 quards (at least one of them must be x);
- two of the three forwards;
- from 4 centers, 1 center.
1C5×2C3×1C4 = 5×3×4 = 60 ways
3. Think about lineups where x is the forward. You must decide.
- 5 quards, 2 guards;
- 2 forwards from 4 forwards (x must be one of them);
- from 4 centers, 1 center.
2C5×1C3×1C4 = 10×3×4 = 120 ways.
Therefore, total number different lineups is - 120+60+120=300.
To learn more about combinations from given link
brainly.com/question/8781187
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