SOLUTION: A committee of 2 tutors and 5 pupils is to be formed among 6 tutors and 10 pupils. In how many ways can this be done, if one particular pupil must be on the committee and two parti

Question 1200763: A committee of 2 tutors and 5 pupils is to be formed among 6 tutors and 10 pupils. In how many ways can this be done, if one particular pupil must be on the committee and two particular pupils must not be on the committee
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

Answer: 525

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Explanation:

Let's calculate how many ways there are to select the 2 tutors.

n = 6 tutors
r = 2 tutor positions on the committee
Order doesn't matter so we use the nCr combination formula.
nCr = (n!)/(r!(n-r)!)
6C2 = (6!)/(2!*(6-2)!)
6C2 = (6!)/(2!*4!)
6C2 = (6*5*4!)/(2!*4!)
6C2 = (6*5)/(2!)
6C2 = (6*5)/(2*1)
6C2 = (30)/(2)
6C2 = 15
There are 15 ways to pick the 2 tutors for the committee.

Put another way:
There are 6*5 = 30 permutations where order matters.
But because order doesn't matter, there are 30/2 = 15 combinations possible.
The dividing by 2 is to correct for the erroneous double-counting.

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Now let's calculate the number of combinations for the students.

Let's say the 10 students have code names A,B,C,...,J

Furthermore, we'll have student A be the student that must have a secured guaranteed seat on the committee.
Students B & C must be left off the committee.
The actual letters picked doesn't really matter since this logic works in general.

Because student A is guaranteed a spot, there are 5-1 = 4 slots left.
The student pool goes from 10 to 10-1-2 = 7
The -1 refers to student A leaving the pool.
The -2 is where we ignore students B and C.

We have
n = 7 students left to pick from
r = 4 slots to fill
Again we use nCr
nCr = (n!)/(r!(n-r)!)
7C4 = (7!)/(4!*(7-4)!)
7C4 = (7!)/(4!*3!)
7C4 = (7*6*5*4!)/(4!*3!)
7C4 = (7*6*5)/(3!)
7C4 = (7*6*5)/(3*2*1)
7C4 = (210)/(6)
7C4 = 35
If you wanted, you can alternatively use Pascal's Triangle to determine these nCr values.

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Wrapping things up:

We found there are 15 ways to select the tutors and 35 ways to select the students.
The order doesn't matter.

Therefore, we have 15*35 = 525 ways to form this committee such that a particular student must be on the committee and 2 other students must be left out.

Side note: Order doesn't matter because each person on the committee has equal rank, and none of the seats are labeled.
If the positions were labeled things like "chairperson", "VP", "treasurer", "secretary", etc, then order would matter.

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