Mathematics, 01.10.2019 21:30 jennsou

Ateacher and 10 students are to be seated along a bench in the bleachers at a basketball game. in how many ways can this be done if the teacher must be seated in the middle and a difficult student must sit to the teachers immediate left?

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Wow !

OK. The line-up on the bench has two "zones" ...

-- One zone, consisting of exactly two people, the teacher and the difficult student.
   Their identities don't change, and their arrangement doesn't change.

-- The other zone, consisting of the other 9 students.
   They can line up in any possible way.

How many ways can you line up 9 students ?

The first one can be any one of 9.   For each of these . . .
The second one can be any one of the remaining 8. For each of these . . .
The third one can be any one of the remaining 7. For each of these . . .
The fourth one can be any one of the remaining 6. For each of these . . .
The fifth one can be any one of the remaining 5. For each of these . . .
The sixth one can be any one of the remaining 4. For each of these . . .
The seventh one can be any one of the remaining 3. For each of these . . .
The eighth one can be either of the remaining 2. For each of these . . .
The ninth one must be the only one remaining student.

   The total number of possible line-ups is

   (9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) = 9! = 362,880 .

But wait ! We're not done yet !

For each possible line-up, the teacher and the difficult student can sit

-- On the left end,
-- Between the 1st and 2nd students in the lineup,
-- Between the 2nd and 3rd students in the lineup,
-- Between the 3rd and 4th students in the lineup,
-- Between the 4th and 5th students in the lineup,
-- Between the 5th and 6th students in the lineup,
-- Between the 6th and 7th students in the lineup,
-- Between the 7th and 8th students in the lineup,
-- Between the 8th and 9th students in the lineup,
-- On the right end.

That's 10 different places to put the teacher and the difficult student,
in EACH possible line-up of the other 9 .

So the total total number of ways to do this is

   (362,880) x (10) = 3,628,800 ways.

If they sit a different way at every game, the class can see a bunch of games
without duplicating their seating arrangement !

Answer from: Quest

step-by-step explanation:

if x = 2 you have to find x to finish the problem

Answer from: Quest

the direct variation says that:

$y \propto x$

then the equation is of the form is given by:

$y = kx$ where k is the constant variation.

from the given table:

consider any value of x and m(x)=y

x = 2 and m(x)=y = 3

substitute these value in [1]; we have;

$3 = 2k$

divide both sides by 2 we get;

$k =\frac{3}{2} = 1.5$

⇒ $m(x) = 1.5x$

check:

put x = -1 and m(x) = -1.5

substitute in the equation:

$m(x) = 1.5x$

$-1.5 = 1.5(-1)$

-1.5 = -1.5 true

therefore, the function m(x) represents a direct variation

graph of this function as shown below:

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Ateacher and 10 students are to be seated along a bench in the bleachers at a basketball game. in ho...

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