States that if there are p ways to do a thing, and q ways to do another thing, and these two things are independent, there are p*q ways to do both things.

Also

The order in which the food and drinks are chosen is not important, which means that the combinations formula is used to solve this question.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

Beverages:

3 from a set of 15. So

$B = C_{15,3} = \frac{15!}{3!12!} = 455$

Appetizers:

3 from a set of 10. So

$A = C_{10,3} = \frac{10!}{3!7!} = 120$

Desserts:

3 from a set of 13. So

$D = C_{13,3} = \frac{13!}{3!10!} = 286$

How many different ways can Luisa pick the food and drinks to serve at the bridal shower?

By the fundamental counting principle, as beverages, appetizers and desserts are independent:

$T = B*A*D = 455*120*286 = 15,615,600$

Luisa can pick the food and drinks to serve at the bridal shower in 15,615,600 different ways.

4 0

4 years ago

Factor completely 12t ^8-75t^4

meriva

$12t^8-75t^4=3t^4\cdot4t^4-3t^4\cdot25=3t^4(4t^4-25)\\\\=3t^4[(2t^2)^2-5^2]\\\\use\ a^2-b^2=(a-b)(a+b)\\\\=3t^4(2t^2-5)(2t^2+5)=3t^4[(t\sqrt2)^2-(\sqrt5)^2](2t^2+5)\\\\=3t^4(t\sqrt2-\sqrt5)(t\sqrt2+\sqrt5)(2t^2+5)$

5 0

3 years ago