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Exercise 6.4 · Q7

Q.In how many ways can a committee of 5 is to be formed from 4 teachers and 6 students so as to include at least 2 students.

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Break the committee-of-5 selection into mutually exclusive cases by the number of students (2,3,4, or 5, since at least 2 students are required and only 4 teachers exist), compute each case with the multiplication principle, then add (addition principle).

[!FORMULA] For each case, ways =6Cs×4C5−s={}^{6}C_{s}\times{}^{4}C_{5-s}, where ss = number of students chosen (from 6 available) and 5−s5-s = number of teachers chosen (from 4 available); total ways = sum over all valid ss.

  1. Committee size is fixed at 5, with at least 2 students. Since only 4 teachers are available, the number of teachers chosen can be at most 4, so students ss can range over s=2,3,4,5s=2,3,4,5.
  2. Case s=2s=2 (2 students, 3 teachers): 6C2×4C3=15×4=60^{6}C_{2}\times{}^{4}C_{3} = 15\times4 = 60.
  3. Case s=3s=3 (3 students, 2 teachers): 6C3×4C2=20×6=120^{6}C_{3}\times{}^{4}C_{2} = 20\times6 = 120.
  4. Case s=4s=4 (4 students, 1 teacher): 6C4×4C1=15×4=60^{6}C_{4}\times{}^{4}C_{1} = 15\times4 = 60.
  5. Case s=5s=5 (5 students, 0 teachers): 6C5×4C0=6×1=6^{6}C_{5}\times{}^{4}C_{0} = 6\times1 = 6. …

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