Mathematics: 8 professors and 11 students are available to serve on committee. (a) How many ways can a committee of four people be selected? (b) How many committees can be formed if each committee must have 3 professors and 4 students?

I think it would be 1.87 that is my answer I m sorry if wrong...

8 professors and 11 students are available to serve

chiefkeef5330

19.01.2022

a) 3876 b) 18480

Step-by-step explanation:

Given , The number of professors = 8

The number of students are available = 11

Total persons = 8+11=19

a) The number of ways to select 4 people = $^{19}C_{4}$

$=\dfrac{19!}{4!(19-4)!}\ \ \ [\because\ ^nC_r=\dfrac{n!}{r!(n-r)!}]\\=\dfrac{19\times18\times17\times16\times15!}{24\times15!}\\=3876$

Hence, the number of ways to select 4 people = 3876.

b) Number of ways to choose 3 professors and 4 students = $^{8}C_{3}\times^{11}C_{4}$

$=\dfrac{8!}{3!(8-3)!}\times\dfrac{11!}{4!(11-4)!}\\\\=56\times330\\\\=18480$

Hence, the numbers of ways to form a committee with 3 professors and 4 students = 18480.

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