A total of 8 representatives need to be chosen from a group of 40 students, of which 24 are female and 16 are male. In how many ways can this be done? What if we require that there be at least 3 female and 3 male representatives?

Probability = 0.0147

Step-by-step explanation:

To find this, we need to know the number of ways the representatives can be picked overall, and then the number of ways 3 females and 3 males can be the representatives. The fraction of the two will be the probability.

The total number of ways the representatives can be picked is the combination of a pool of 40 students and choosing 8 is; C(40,8) = 40!/((8!)(40 - 12)!)

= 76904685

We have gotten the number of ways the representatives can be picked.

Now let's find how many ways can the jury consist of 3 females and 3 males

For the males, there are 16 in the pool and we're picking 3:

Thus, C(16,3) = 16!/((3!)(16 - 3)!) = 560

For the females, there are 24 in the pool and we're picking 3:

Thus, C(24,3) = 24!/((3!)(24 - 3)!) = 2024

Thus, number of men and women choices = 2024 x 560 = 1133440

So; probability = 1133440/76904685 = 0.0147

3

step-by-step explanation:

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)