Question

Suppose you are forming a committee from a group of 15 biology student, 12 math students, and 9 physics students. How many possibilities are there if

Answer 1

12,474,000 possibilities

Explanation:

The question is incomplete. Here is the complete question.

Suppose you are forming a committee from a group of 15 biology student, 12 math students, and 9 physics students. How many possibilities are there if your committee needs to have at most 2 biology students, exactly 3 math students, and exactly 2 physics students?

To tackle this question, we will use the concept of combination since it deals with selection. Generally, selecting 'r' objects out of 'n' pools of object can be done using the formula;

nCr = n!/(n-r)!r!

If we are to form a committee of at most 2 biology students, exactly 3 math students and exactly 2 physics students from a group of 15 biology student, 12 math students, and 9 physics students, this can be done in the following ways;

For Physics students:

Selecting exactly 2 physics students from a group of 9 students will be:

9C2 =
`(9!)/((9-2)!2!)\\`

=
`(9!)/((7)!2!)\\`

`= (9*8*7!!)/((7)!2!)\\= 9*4\\= 36ways`

for Mathematics students:

Selecting exactly 3 math students from a group of 12 students will be:

`12C3 = (12!)/(12-3)!3!)\\= (12!)/(9!3!)\\= (12*11*10*9!)/(9!*6)\\= 220 ways`

For Biology Students:

Selecting at most 2 biology students from a group of 15biology student will be:

15C1 * 15C2 (at most 2 students)

=
`(15!)/(14!1!) * (15!)/(13!2!)\\\\`

= 15*105

= 1,575 ways

The total number of possibilities will be = 36*220*1,575 = 12,474,000 possibilities