The smith family has 4 sons and 3 daughters. in how many ways can they be seated in a row of 7 chairs such that all 3 girls sit next to each other? answer asap! i will give out a brainiest for the 1st correct answer.

Step-by-step explanation:

Given that,

Smith family

Sons = 4

Daughters = 3

Let D represents the daughter

Let S- represents the sons

DDD

So, we want to arrange the children in a seven chair row, such that all the daughters are sitting together.

The children are not identical so we have the arrangement below

e.g D¹D²D³S¹S²S³S⁴

D¹ represents first daughter

D² represents second daughter

D³ represents third daughter

S¹ represents first son

S² represents second son

S³ represents third son

S⁴ represents fourth son

If we take the daughters as a one entity, I.e. we will see all the three daughters as just one D.

Let the three daughters represent X

Then, we have XS¹S²S³S⁴

The sons are not identical, so they switch positions

So, arranging this is

5! = 5×4×3×2 × 1 = 120ways

Now, we will assume that the daughters are not identical too, so they can be arrange in 3! ways

D¹D²D³

3! = 3 × 2 × 1 = 6 ways

Then, the total arrangement is

6 × 120 = 720 ways

So, they can be arrange in 720ways

it could be a ..idk tho

correct answer is infinitely many solutions

step-by-step explanation:

2(2x + 3) = -4 + 2(2x + 5)

solving the above equation:

4x + 6 = -4 +4x +10

combining like terms

4x -4x = -4+10-6

0=0

since both sides are 0=0, we end up into true statement it means it has infinite many solutions.

so correct answer is infinitely many solutions