SOLUTION: How many different committees can be formed from 7 teachers and 43 students if the committee consists of 3 teachers and 3 ​students?

Algebra ->  Permutations -> SOLUTION: How many different committees can be formed from 7 teachers and 43 students if the committee consists of 3 teachers and 3 ​students?      Log On


   

Question 1201328: How many different committees can be formed from 7 teachers and 43 students if the committee consists of 3 teachers and 3 ​students?
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52817) About Me  (Show Source):
You can put this solution on YOUR website!
.

C%5B7%5D%5E3.C%5B43%5D%5E3 = 35*12341 = 431935.    ANSWER

Solved.


Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

Answer: 431,935
This number is slightly less than 432 thousand


Explanation:


There are 7 teachers to choose from for slot1.
There are 6 teachers to choose from for slot2.
There are 5 teachers to choose from for slot3.
We count down by 1 each time we need to fill another slot.
This is because we cannot reselect a certain teacher more than once.

There would be 7*6*5 = 42*5 = 210 permutations if the order of teachers mattered.
However, the order doesn't matter on a committee.
Each member has equal rank, and the seats aren't labeled.
If we had seat names like "chair person", "VP", "secretary", "treasurer", etc, then order would matter.

Let's look at the number of ways to arrange a group of 3 people.
I'll call them A,B,C
We have 3*2*1 = 6 ways to do so as listed below.
  1. ABC
  2. ACB
  3. BAC
  4. BCA
  5. CAB
  6. CBA
Because there are 6 ways to rearrange a group of three people, we must divide the permutations result we got earlier by 6. This will give us how many combinations of teachers there are.

210/6 = 35
This corrects the over-counting done by a factor of 6.

There are 35 ways to select the three teachers from a pool of seven. Order does not matter.

Here is another way to get the "35"
n = 7 teachers
r = 3 selections
Use the nCr combination formula since order doesn't matter
n C r = (n!)/(r!(n-r)!)
7 C 3 = (7!)/(3!*(7-3)!)
7 C 3 = (7!)/(3!*4!)
7 C 3 = (7*6*5*4!)/(3!*4!)
7 C 3 = (7*6*5)/(3!)
7 C 3 = (7*6*5)/(3*2*1)
7 C 3 = 210/6
7 C 3 = 35
Note: we have 7*6*5 = 210 up top and 3*2*1 = 6 down below
These values were mentioned in the previous section.

Yet another way to get the 35 is to look at Pascal's Triangle
Look at the row that has 1,7,... at the start
The start index is r = 0 which corresponds to the "1" at the left edge.
r = 1 then corresponds to the 7 and so on
Count over four spaces to arrive at the index r = 3, and this value is the first copy of "35"

---------------------

Unfortunately Pascal's triangle is not practical when determining how many ways there are to pick the three students from a pool of 43.
It can be done, but the triangle would be really massive.

So I'll use the nCr formula
n = 43 students
r = 3 student selections
n C r = (n!)/(r!(n-r)!)
43 C 3 = (43!)/(3!*(43-3)!)
43 C 3 = (43!)/(3!*40!)
43 C 3 = (43*42*41*40!)/(3!*40!)
43 C 3 = (43*42*41)/(3!)
43 C 3 = (43*42*41)/(3*2*1)
43 C 3 = 74046/6
43 C 3 = 12341

Or a shortcut
43*42*41 = 74046 permutations
74046/6 = 12341 combinations
We divide by 6 for the same reasoning as applied to the teachers.

---------------------

Let's recap
  • 35 ways to pick the three teachers (order doesn't matter).
  • 12341 ways to pick the three students (order doesn't matter).
Therefore, we have 35*12341 = 431,935 ways to pick the committee of 3 teachers and 3 students (order doesn't matter).