A committee must be formed with 4 teachers and 5 students. If there are 11 teachers
to choose from, and 11 students, how many different ways could the committee be
made?

Respuesta :

Answer:

11C4 × 11C5

[tex] \binom{11}{4} \times \binom{11}{5} = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} \times \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = 330 \times 462 = 152460[/tex]