Question

How many different committees can be formed from 11 teachers and 41 students if the committee consists of 4 teachers and 3 students?

In how many ways can the committee of 7 members be selected?

Answer 1

how do I solve for 11C4? and 41C3?

Answer 2

???

Answer 3

The number of combinations of teachers selected 4 at a time is 11C4.
The number of combinations of students selected 3 at a time is 41C3
The number of ways of selecting the committee of 7 is given by:
\[\frac{11!}{4!7!}\times \frac{41!}{3!38!}=\frac{11\times 10\times 9\times 8}{4\times 3\times 2\times 1}\times\frac{41\times40\times 39}{3\times 2\times 1}=you\ can\ calculate\]

Answer 4

\[\Large nCr = \frac{ n! }{ r! (n-r)! }\]
so
\[\Large 11C4 = \frac{ 11! }{ 4! (11-4)! }\]