Question

How many different committees can be formed from 12 teachers and 43 students if the committee consists of 3 teachers and 4 ​students?

The committee of 7 members can be selected in BLANK
different ways.​

Answer 1

To determine the number of different committees that can be formed from 12 teachers and 43 students with 3 teachers and 4 students on the committee, calculate the combinations separately and multiply them. The number of different committees is 27,150,200.

Step-by-step explanation:

To form a committee of 3 teachers out of 12, we need to use combinations since the order of selection does not matter. The number of ways to choose 3 teachers is given by the combination formula C(n, k) = n! / (k! (n - k)!), where n is the total number of items, and k is the number of items to choose.

For the teachers, the calculation is C(12, 3) = 12! / (3! (12 - 3)!) = (12 × 11 × 10) / (3 × 2 × 1) = 220 ways.

Similarly, to select 4 students out of 43, we use the combination formula as well: C(43, 4) = 43! / (4! (43 - 4)!) = (43 × 42 × 41 × 40) / (4 × 3 × 2 × 1) = 123,410 ways.

The total number of different committees that can be formed is found by multiplying the two results together: 220 × 123,410 = 27,150,200 different committees.

Therefore, a committee of 7 members consisting of 3 teachers and 4 students can be selected in 27,150,200 different ways.

Answer 2

`\displaystyle 27150200`

Step-by-step explanation:

we are two conditions

• committees can be formed from 12 teachers and 43 students
• the committee consists of 3 teachers and 4 students

In choosing a committee, order doesn't matter; in case of teachers we need the number of combinations of 3 people chosen from 12

remember that,

`\displaystyle\binom{n}{r} = (n!)/(r!(n - r)!)`

with the condition we obtain that,

• 
  `n = 12`
• 
  `r = 3`

therefore substitute:

`\displaystyle\binom{12}{3} = (12!)/(3!(12 - 3)!)`

simplify Parentheses:

`\displaystyle\binom{12}{3} = (12!)/(3! \cdot9!)`

rewrite:

`\rm \displaystyle\binom{12}{3} = (12 * 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1)/((1 * 2 * 3 )\cdot1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9)`

reduce fraction:

`\rm \displaystyle\binom{12}{3} = (12 * 11 * 10)/(1 * 2 * 3 )`

rewrite 12 and 10:

`\rm \displaystyle\binom{12}{3} = (3 * 2 * 2 * 11 * 10)/(1 * 2 * 3 )`

reduce fraction:

`\rm \displaystyle\binom{12}{3} = 2 * 11 * 10`

simplify multiplication:

`\rm \displaystyle\binom{12}{3} = 220`

In case of students we need the number of combinations of 4 students choosen from 43 therefore,

`\displaystyle\binom{43}{4} = (43!)/(4!(43 - 4)!)`

simplify which yields:

`\displaystyle\binom{43}{4} = 123410`

hence,

The committee of 7 members can be selected in BLANK different ways is

`\displaystyle 123410 * 220`

`\displaystyle \boxed{27150200}`

and we're done!