Sid intended to type a seven-digit number, but the two 3's he meant to type did not appear. what appeared instead was the five-digit number 52115. how many different seven-digit numbers could sid have meant to type? a. 10b. 16c. 21d. 24e. 27

D. 21ways.

Step-by-step explanation:

This is based on selection, fixing two '3' numbers into 5numbers to make it 7 numbers can be done in 7C2 ways.

Generally, nCr = n!/(n-r)!r!

7C2 = 7!/(7-2)!2!

=7!/5!2!

=7×6×5!/5!×2!

=7×6/2

=21ways (D)

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)