Two students from a class of 20 are chosen at random to escort a guest speaker to a presentation. The first student will introduce the speaker. The second will thank the speaker when she is done. In how many ways can the two students be chosen?

380 different ways

Step-by-step explanation:

We can solve this problem using a permutation formula, as the order in a group of two students matter.

So, for this problem, we have 20 students and want to form groups of 2, so we have a permutation of 20 choose 2:

P(20,2) = 20! / (20-2)! = 20! / 18! = 20 * 19 = 380 different ways

Another way to solve this problem is:

For the first student being chose, we have 20 students available, and for the second student, we have just 19, as one was already chose, so the number of ways to choose these two students is 20 * 19 = 380

what's the question?

step-by-step explanation:

there is no question.

hello from mrbilldoesmath!

answer:   (7x + 6y) ( 49x^2 - 42xy + 36y^2)

discussion:

in general, the factorization of a^3 + b^3 is

a^3 + b^ 3 = (a + b) ( a^2 - ab + b^2)

in our case note that 343 = 7^3 and 216 = 6^3 so the equation becomes:

343x^3 + 216y^3 = (7x)^3 + (6y)^3

if you think of a (above) as 7x and b as 6y the factorization becomes

343x^3 + 216y^3 =

((7x + 6y) (   (7x)^2 - 7x*6y + (6y)^2)   =

(7x + 6y) ( 49x^2 - 42xy + 36y^2)

you,

mrb

(a + b) (a^2 - a b + b^2)