Three boys and six girls are being seated in a row of nine chairs on a stage which are numbered from left to right. How many seating arrangements are there if: Girls sit in the first two seats as well as the last seat?

Step-by-step explanation:

Given that,

We have 3 boys and 6 girls

They are sitted in a row of 9 chairs.

Give that that three girls are position in the the first two seat and last seat

Then the arrangement of the girls in the first two rows and last seat

6P3= 6!/(6-3)!

6P3=6!/3!3!

6P3=6×5×4×3!/3!

6P3=120ways

This is the arrangement of the girls in the first two seats and last seat

Now, there are six seat space, where 3boys and 3girls can seat.

i.e we want to arrange

BBBGGG in the remaining six spaces

The arrangement become

6!/3!3!

6×5×4×3!/(3!×3×2×1)

=20ways

But their are other 3 girls that can still be arrange in that postion, so the total postion is

3×20ways=60ways

Total arrangement becomes

120+60=180ways

27

step-by-step explanation:

(2x + 16) + (x) = 180

step-by-step explanation:

given,

abcd is a quadrilateral inscribed in a circle

angle a is labeled x plus 16, (x + 16)

angle b is labeled x degrees, (x)

angle c is labeled 6x minus 4, (6x – 4)

angle d is labeled 2x plus 16, (2x + 16)

what equation can be used to solve for angle b?

we know that the sum of the opposite angles of a simple quadrilateral inscribed in a circle is 180 degrees.

angle b is opposite angle d

therefore, angle b + angle d = 180

(x) + (2x + 16) = 180 

same as (2x + 16) + (x) = 180