Question

How many ways can an investigation committee be formed from a set of 5 teacher and 7 student leaders if a committee needs 2 teacher and 3 student leaders?
plss help me​

Answer 1

Answer:

350

Step-by-step explanation:

first to form the committee we need 2 teachers from 5 so select 2 from 5 which is 5C2 and for student leader we need 3 out of 7 which is 7C3.

answer= 5C2*7C3

           =10*35

           =350

Answer 2

Answer:

350 ways the committee can be formed with 2 teachers 3 students.

Step-by-step explanation:

A combination is a selection of items from a set that has distinct members, such that

$n_{c}_{r} =$ $\frac{n!}{r!(n-r)!}$

$n_{c} _{r}$ = number of combinations

$n$ = total number of objects in the set

$r$ = number of choosing objects from the set

Given that there are 5 teachers and 7 students

total 5 + 7 =12

A committee of 5 members to be formed with 2 teachers and 3 students.

The number of ways the committee can be formed

= $5_{c} _{2}$ × $7_{c} _{3}$

=$\frac{5!}{2!*3!}$ × $\frac{7!}{3!*4!}$

=$\frac{(5)(4)}{(2)(1)}$ ×  $\frac{(7)(6)(5)}{(3)(2)(1)}$

= 10 × 35

=350