6 Teachers and 12 students volunteer for a committee to discuss school activities. How many committees of 5 people can be made if:
a) there are no restrictions
b) there must be 3 students on the committee
c) there must be at least one teacher on the committee
d) Todd and Mr Ryan cannot be on the committee together.
a) This is equal to 18 choose 5, which is 8568.
b) This equals 12 choose 3 plus 6 choose 2, which is 220 + 20 = 240.
c) This equals 6 choose 1 plus 12 choose 5, which is 6 + 792 = 798
d) This equals 8568 minus the number of groups in which Todd and Mr Ryan are together. There are 16 choose 4, or 1820, possible groups with both Todd and Mr. Ryan, so the number of groups where that is not true is 8568 - 1820 = 6748.
I’m fairly sure these are all correct. If not, someone please correct me.
thank you so much @Hermit9 . I forgot that I had to do choose as order does not matter.
No problem!
the only thing is in b) 6 choose 2= 15 I believe
Also for c) the first part is 6 choose 1 but the next equation would be 17 choose 4 because it says at least 1 teacher so the other 4 people do not matter if they are teachers are students
Thanks, you’re right.
@MITer94 Could you double check the answers? Thank you so much for all your help. You don’t know how much it means to me.
a) 18 c 5= 8568 different committees
b)(12 c 3) + (6 c 2)
= 220 +15
=235 different committees
c)6 c 1 + 17 c 4
=6 + 6188
=6194
d) i’m still not 100% sure how to do it
a) is correct.
b) You should multiply here. 12C3 = 220 ways to pick 3 students, and you need to pick 2 teachers, i.e. 6C2 = 15 ways to pick teachers. 220*15 = 3300 (see [here](https://en.wikipedia.org/wiki/Rule_of_product)). I’m also assuming the problem means exactly 3 students.
c) Neither you nor @Hermit9 got it. This one is slightly trickier.
Note the problem states we need at least one teacher. In simple English terms, it means we can have 1, 2, 3, 4, 5, or 6 teachers. You can compute for each value of n, where n = # teachers, but a much easier way is to count the number of committee with no teachers and subtract it from 18C5 (the total number of possible committees). This method is known as [complementary counting](Art of Problem Solving).
ways = 18C5 - (# committees with no teacher)
= 18C5 - 12C5 (note: a committee with no teacher must consist of all students)
= 7776
d) This problem can also be done using complementary counting. Hermit9’s solution is correct except that there are 16C3 (not 16C4) committees with both Todd and Mr. Ryan.
thank you again @MITer94