a teacher 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?

Answers

Answer 1

Answer:
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 2

Answer:

Step-by-step explanation:

362,880

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Which number can be added to a rational number to explain that the sum of rational number and a irrational number is irrational?A) -8
B) 2 1/6
C) 5
D) 27

Answers

Answer:

Step-by-step explanation:

is correct. 5

cannot be expressed as a fraction. A rational number is any number that can be expressed as the quotient or fraction p/q of two integers, with the denominator q not equal to zero.

I am not sure this is right, but I would say A

Jess observed that 60% of the people at a mall on a particular day shopped for clothes. If 2500 people at the mall did not shop for clothes that day, the number of people who shopped for clothes that day was ______. (only put numeric values, no other symbols)

Answers

The 2500 people that did not shop for clothes that day accounts for 40% of the total people present in the mall. The total number of people present at the mall that day is the quotient obtained when 2500 is divided by 40% which is 6250. Since the number of people who shopped for clothes that day is 60% of the total number of people, multiply 6250 by 60%. This gives an answer of 3750.

Therefore, there are 3750 people who shopped for clothes that day.

Answer:

3750

Step-by-step explanation:

A ride on the roller coaster costs 4 tickets while the boat ride only costs 3 tickets. Michael went on the two rides a total of 10 times and spent a total of 37 tickets.

Answers

let x = roller coaster ticket
let y = boat ride ticket

x + y = 10
4x + 3y = 37

x = 10 - y
4x + 3y = 37
4(10-y) + 3y = 37
40 - 4y + 3y = 37
-y = 37 - 40
-y = -3
y = -3/-1
y = 3

x = 10 - y
x = 10 - 3
x = 7

x = 7 ; y = 3

4x + 3y = 37
4(7) + 3(3) = 37
28 + 9 = 37
37 = 37

Answer:

D. 4r + 3(10 - r) = 37

Step-by-step explanation:

worked on edge 2022

3 cards are drawn at random from a standard deck. Find the probability that all the cards are hearts.

Find the probability that all the cards are face cards.

Note: Face cards are kings, queens, and jacks.

Find the probability that all the cards are even.

(Consider aces to be 1, jacks to be 11, queens to be 12, and kings to be 13)

Answers

Answer:

i dont really understand what you mean it says 3 cards are drawn and it says to find the probability that the cards are all hearts but what is the question asking how do we find the probability

Step-by-step explanation:

Final answer:

The probabilities of drawing all hearts, face cards, or even cards are calculated with the formula: $(n/52) * ((n-1)/51) * ((n-2)/50)$ where n is the total number of cards that match the desired outcome.

Explanation:

The subject here is probability, specifically, how to determine the likelihood of a particular outcome when drawing cards from a standard deck. Let's deal with each probability one at a time.

The probability that all the cards are hearts: There are 13 hearts in a deck of 52 cards. So the probability that the first card is a heart is 13/52, the second is 12/51 (because one heart is already drawn), and the third is 11/50. So, the overall probability is $(13/52) * (12/51) * (11/50).$
The probability that all the cards are face cards: There are 12 face cards (kings, queens, and jacks) in a deck. Using the same principle, the probability is $(12/52) * (11/51) * (10/50).$
The probability that all the cards are even: The 'even' cards are 2, 4, 6, 8, 10, which have 4 of each (hearts, diamonds, spades, clubs) totaling 20 cards. So, the probability is $(20/52) * (19/51) * (18/50).$

Learn more about Probability here:

brainly.com/question/32117953

#SPJ3

What is the recursive equation for the following sequence: 2,5,8,11

Answers

f(n)=f(n-1)+3 such that
f(1)=2

Can yall help meee please

Answers

Answer:

$jupiters \: mass \: is \: approximatelty \: \n \boxed{{6 * 10}^(3) \: kg} \: times, \: \n more \: than \: mercurys \: mass$

Step-by-step explanation:

$............................................................... \n in \: other \: to \: get \: right \: ans \to \n you \: divide \: mass \: of \: jupiter \: by \: the \: mass \n \: of \: mercury : \: so \to \n ............................................................... \n jupiters \: mass = 1.898 { * 10}^(27) \: kg \n mercurys \: mass =3.3 * {10}^(23) \: kg \n their \: mass \: ratio \: is : \n = \frac{1.898 { * 10}^(27)}{3.3 * {10}^(23)} = 0.5751515152 * {10}^(4) = \n \boxed{5,751.515152} \n hence \: jupiters \: mass \: is \: approximatelty \: \n {6 * 10}^(3) \: times \: more \: than \: mercurys \: mass$

6x103 because it explains that it’s 6 times but 10 in 3 times.

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