6 girls and 2 boys are to be seated in a row. Find the number of ways that can be done if 2 boy must have exactly 4 girls seated between them.

Hello!

6 girls and 2 boys are to be seated in a row, with 4 girls seated btwn the 2 boys.

Now let us number 8 seats, starting from the left as follows:

S1, S2, S3, S4, S5, S6, S7, S8

Now the following cases arise regarding the seating arrangement of the boys -

B1 seated on S1 and B2 seated on S6 ( leaving S 2,3,4 & 5)

Now in the remaining 6 seats, 6 girls can be arranged in 6! ways = 720 ways.

On interchanging B1 and B2′s place, we get 720 more arrangement

Therefore, current total = 1440 ways

2. B1 - S2 and B2 - S7

1440 more arrangements can be formed as in previous case.

3. B1 - S3 and B2 - S8

1440 more arrangements can be formed as in previous case.

Therefore total ways to arrange the seating plan are 1440 × 3 = 4320 ways

Hope This helped you! :D

step-by-step explanation:

5x + 5

step-by-step explanation:

you should always add the diagrams and graphs we cannot see. this particular question is kind of borderline because we can make a graph for you.

(f + g)x = -x^2 +3x + 5 + x^2 + 2x

(f + g)x = 5x + 5

the graph of z = 5x + 5 is shown below. always use desmos when in doubt.