In how many ways can a photographer at a wedding arrange 6 people in a row from a group of 10 people, where the bride and the groom are among these 10 people, if a) the bride must be in the picture? b) both the bride and groom must be in the picture? c) exactly one of the bride and the groom is in the picture?

(a) 126 ways

(b) 70 ways

(c) 140 ways

Step-by-step explanation:

(a) if the bride must be in the picture.

Now, if the bride must be in the picture, it means we are selecting 5 out of 9 people.

This is a combination problem and can be solved as 9C5 = 126 ways

(b) if both the bride and the groom must be in the picture, it means a already have 2 places taken. What is left is to select 4 people from 8 people.

That is 8C4 = 70 ways

(c) Exactly one of the bride and the groom.

Here if the wife is in the picture, this means the husband cannot be in the picture. It simply means we are selecting 4 out of 8 people.

This is same as (b) above and it gives 70 ways.

Now If the husband is in the group, the wife cannot be there. We simply do same and this gives 70 ways. The total number of arrangements now yield 70 × 2 = 140 ways.

step-by-step explanation:

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3