John is painting a picture. He has green, red, yellow, purple, orange, and blue paint. He wants to use four different colors on this picture. If order doesn't matter, how many ways can John pick four colors for his picture, if one of them is green?

a) .44
b) .52
c.) .23
d) .35

I'm gonna say maybe 10 yah 10

here's why

In can pick in 10 ways.

Solution:

John has green, red, yellow, purple, orange, and blue paint = total 6 colours

he wants to have 4 different colours in the painting.

But in those 4 colors green is must, it should always be there.

So except green we have total 5 colours (red, yellow, purple, orange, and blue) from which we can pick any 3 colors(as green will be one in four colours).

Then we can choose 3 colours from 5 colours in x ways.  

x ways = = \left(\frac{5 !}{3 !(5-3) !}\right)=(3!(5−3)!5!)

=\left(\frac{5 !}{3 ! \times 2 !}\right)=(3!×2!5!)

=\frac{5 \times 4 \times 3 !}{3 ! \times 2 !}=3!×2!5×4×3!

=\frac{5 \times 4}{2}=25×4

=\frac{20}{2} = 10=220=10

step-by-step explanation:

you divide the denominator by the numerator on both sides and if the quotients are equal then the fractions are equivalent

cheater

step-by-step explanation: