A committee must be formed with 4 teachers and 5 students. If there are 10 teachers to choose from, and B students, how many different ways could the committee be
made?

We have to choose 5 teachers from 9 teachers.

So, number of ways doing this is (95)

Similarly for. Number of ways choosing 4 students from 15 are (154)

Notice that we have to choose both that is teachers as well as students.

So it is like a case of intersection as of set which is teachers and students

So,total number of ways = (154) × (95)

⟹  15!(15–4)! × 4!  ×  9!(9–5)! × 5!

⟹  1365  ×  126

⟹  171990 ways

Step-by-step explanation:

huh? im pretty sure it is in 199

step-by-step explanation:

a) is $10.34

b) is 6 toppings