Question 949746
This is just my interpretation. Judge for yourself, because I may be overlooking some mistake.
 
There are lots of ways to make a committee of 18 students,
if you have {{{7+6+7+4=24}}} eligible candidates.
In fact there are
{{{(matrix(2,1,24,18))=(matrix(2,1,24,6))=24*23*22*21*20*19/(1*2*3*4*5*6)=23*22*19*7*2=134596}}}
possible groups of 18 people that can be selected from a pool 24 people,
or from a different point of view,
there are {{{134596}}} possible groups of 6 rejected candidates that you can send home.
 
However, some of those {{{134596}}} possible committees will fail the requirement to consist of at least 1 freshmen, 1 sophomore, 2 juniors, and 2 seniors.
How many of the {{{134596}}} possible committees fail the requirement, and how/why do they fail?
 
No committee can fail because of not including 1 freshman,
because without the 7 freshmen, we have only
{{{24-7=17}}} eligible candidates, and that is not enough for an 18-member committee.
 
Can a committee of 18 fail to include 1 sophomore?
Yes, because we have {{{24-6=18 non-sophomores,
and they would form {{{1}}} of the {{{134596}}} possible committees.
We cannot count that a {{{1}}} committee.
 
Can a committee of 18 fail to include 2 juniors?
Yes, while it is impossible to exclude all 7 juniors,
and form the committee with the {{{24-7=17}}} remaining candidates,
we can send home 6 juniors,
and still have {{{24-6=18}}} members for our committee.
We can do that 7 different ways,
keeping a different one of the 7 juniors for the committee.
That way we could make {{{7}}} different committees
that would fail to have the requisite 2 juniors.
We cannot count those {{{7}}} committees, either.
 
Can a committee of 18 fail to include 2 seniors?
Excluding all 4 seniors,
we would still have {{{24-4=20}}} students for our committee.
With those 20 students, we could make
{{{(matrix(2,1,20,18))=(matrix(2,1,20,2))=20*19/(1*2)=190}}}
different committees that fail the 2-senior requirement,
because they have no senior.
 
We could also send 3 seniors home,
and have {{{24-3=21}}} students left to form an 18-member committee.
that would fail the requirement to have 2 seniors, because it would have only one senior.
We would have 4 choices as to what senior we would keep,
and for each of those choices,
we would still have to pick 17 more students from the 20 non-seniors.
We would have {{{(matrix(2,1,20,17))=(matrix(2,1,20,3))=20*19*18/(1*2*3)=1140}}} possible groups of 17 non-seniors,
that combined with 4 different choices of one senior would make
{{{4*1140=4560}}} committees with only 1 freshman.
 
So, we have to exclude from the total of {{{134596}}} possible committees
the {{{1}}} committee with no sophomore,
the 7 committees with only 1 junior,
the {{{190}}} committees with no senior, and
the {{{4560}}} committees with only 1 freshman.
Counting them all, that is {{{1+7+190+4560=4758}}} committees that fail the requirement one way or another.
Could there be one or more committees counted twice in those 4758, because they fail to meet the requirement more than one way?
The committee that is missing a sophomore includes all the non-sophomores, so it cannot fail by missing any other requirement,
and the committees that have only 1 junior, must include all non-juniors,, so it cannot fail by missing any other requirement.
As for the seniors, a committee with only one, and a committee with none cannot be the same, and I think I am counting them properly.
 
Subtracting those {{{1+7+190+4560=4758}}} committees that fail the requirement,
we are left with
{{{134596-4758=highlight(1298388)}}} committees that meet the requirement.
 
NOTE: The folks at the forum in the artofproblemsolving website would probably instantly answer this question, but they may be too succinct with the explanation, or may feel insulted if it is asked in a forum at a level that considers it too easy a problem;