How to solve this issue using basics of counting? We have 20 tutors in a mathematics college, 8 are specialized in algebra, 5 are specialized in discrete mathematics, 4 are specialized in applied mathematics and 3 are specialized in statistic.

How many ways as we can form a scientific committee consisted of 8 tutors in each of these situations:


It has 2 tutors from each section.
Three of them at least must be from discrete mathematics section.
One of them at least must be from algebra section.
Three of them must be from discrete math, but we have to take into consideration that 1 tutor from discrete math and 1 tutor from statistic mustn't be together in the committee.


Please explain your solutions.
You need to choose two (combinations - order doesn't matter) from each group, for a total of choosing 8.

Choose 2 from the 8 alebra, 2 from the 5 discrete math, 2 from the 4 applied math, and 2 from the 3 stats:

$8 \choose  2$$\times$ $5 \choose  2$ $\times$ $4 \choose  2$$\times$   $3 \choose  2$ = 5 040
You have to choose at least three from discrete math. That means that you have to choose 3, 4, or 5 (there are a maximum of 5 that can be chosen).

So, combinations for choosing three from discrete math is the 5 discrete math, choose 3, multiplied by what's left - the other 15 tutors, choose 5 to make 8 total chosen:

$5 \choose 3$ $\times$ $15 \choose 5$

Similarly for choosing four from discrete math, choose the 4 of 5 from discrete math, then choose the remaining 4 tutors to be chosen from the other 15 tutors:

$5 \choose 4$ $\times$ $15 \choose 4$

Then for choosing all 5 discrete math tutors:

$5 \choose 5$ $\times$ $15 \choose 3$

Now you have all the combinations for choosing 3, 4 and 5 discrete math tutors. Just add them all up to get the final answer of choosing at least three:

$5 \choose 3$ $\times$ $15 \choose 5$ +  $5 \choose 4$ $\times$ $15 \choose 4$ + $5 \choose 5$ $\times$ $15 \choose 3$ = 37 310
At least one of them must be from algebra. We can use the compliment here. At least one of them being from algebra is the same as NOT zero of them being from algebra. We can calculate what we don't want and then subtract that from the sample space (total number of all possible combinations) to get what we do want.

Sample space (20 tutors, choose 8 of them): n(s) = $20 \choose 8$

Then, the combinations to get ZERO tutors from algebra is 8 from algebra, choose 0 of them, multiplied the rest of the tutors, choose the 8 remaining chooses:

$8 \choose 0$ $\times$ $12 \choose 8$

And then finally, just subtract that from the sample space:

n(s) - $8 \choose 0$ $\times$ $12 \choose 8$ = $20 \choose 8$ - $8 \choose 0$ $\times$ $12 \choose 8$
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;= $20 \choose 8$ - $12 \choose 8$
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;= 125 475
Exactly 3 have to be chosen from discrete math, but one person from discrete math and one person from stats don't get along, so they can't be together.

So, let's just exclude the bad egg from discrete math:

5 tutors from discrete math, but take out 1, so it's 4, choose 3. Then multiply that by the 15 other tutors that are not from discrete math and choose the remaining 5 chooses:

$4 \choose 3$ $\times$ $15 \choose 5$

That, OR, exclude the bad egg from stats:

Choose 3 from all 5 in discrete math, but choose the remaining 5 chooses from only 14 of the remaining tutors, because you have to exclude the bad egg from the stats:

$5 \choose 3$ $\times$ $14 \choose 5$

And then you would just add those up:

$4 \choose 3$ $\times$ $15 \choose 5$ + $5 \choose 3$ $\times$ $14 \choose 5$ = 32 032




On number 4, I'm not so sure my answer is correct. The way I did it, the combinations of neither of them being on the committee may have overlapped, so you would have to subtract that from the answer, like this:

32 032 - $4 \choose 3$ $\times$ $14 \choose 5$

And the 14 choose 5 may be wrong as well. You may have to replace $14 \choose 5$ with:

$3 \choose 2$ $\times$ $12 \choose 3$ + $3 \choose 1$ $\times$ $12 \choose 4$ + $3 \choose 0$ $\times$ $12 \choose 5$

because 3 stats choose only 2 (excluding bad egg) and then choosing the other 3 from the remaining 12 tutors (not stats or discrete) and so on down to choosing zero from the stats.

I can't think right now, so someone else has to confirm this for me.
You need 8 tutors and must have 2 tutors from each group.

From algebra you need to select 2 tutor out of 8  so it can be done in $8\choose 2$ ways.

Similarly for discrete mathematics it can be done in $5\choose2$ ways.

For applied mathematics it can be done in $4\choose3$ ways.

And for statistics it can be done in $3\choose2$ ways.

Since each of the selections made for the committee are independent so all the above ways will be added as:

$8\choose2$+$5\choose2$+$4\choose2$+$3\choose2$=47 

Hence committee can be formed in 47 ways.
Out of 8 members you need at least 3 from discrete mathematics.

It means that for selecting 3 from discrete mathematics you have $8\choose3$ ways.

Now for selecting other 5 members you have no conditions(means they cam be from any group).

Now since you have already selected 3 members so now you are left with only 17 members. 

So for rest of the 5 members you can make selected in $17\choose5$ ways.

Now total number of ways would be $8\choose3$+$17\choose5$=6244 ways
Out of 8 members you have to select at least 1 from algebra.

It means you have $8\choose1$ ways of selecting tutor from algebra.

Now for selecting other 7 members you have no conditions(means they cam be from any group).

Now since you have already selected 1 members so now you are left with only 19 members. 

So for rest of the 7 members you can make selected in $19\choose7$ ways.

Now total number of ways would be $8\choose1$+$19\choose7$=50396 ways.