Question 1206993: How many different committees can be formed from 6 teachers and 39 students if the committee consist of 3 teachers and 3 students?
Found 2 solutions by MathLover1, math_tutor2020: Answer by MathLover1(20849) (Show Source): Answer by math_tutor2020(3817) (Show Source):
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6 teachers
3 slots for teachers
6 choices for the first teacher slot
5 choices for the second teacher slot
4 choices for the third teacher slot
6*5*4 = 30*4 = 120 permutations for the teacher portion of the committee.
If order mattered then we wouldn't adjust this value.
However, order doesn't matter since we're dealing with a committee that has unnamed seats.
Each person is of equal rank.
A committee like {A,B,C} is the same as {B,A,C}.
If there were named seats like "president", "VP", etc then order would matter.
Since order doesn't matter we divide by 6 which is the number of ways to arrange any trio of people (because 3*2*1 = 6)
Therefore, we have 120/6 = 20 different teacher combinations.
Note this value can be found in Pascal's Triangle in the row that starts with "1,6,..."
Through similar steps you should find that there are (39*38*37)/(3*2*1) = 9139 different student trios. Again order doesn't matter.
You can reach this value through the nCr combination formula with n = 39 and r = 3.
Pascal's Triangle is NOT recommended here since n is fairly large. The triangle would be very massive.
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Let's wrap things up
20 different teacher trios
9139 different student trios
Order does not matter.
Multiply the values to get the answer
20*9139 = 182780
This value is slightly less than 183 thousand.
Answer: 182780
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