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Rus_ich [418]
3 years ago
14

You have 6 reindeer, Prancer, Rudy, Balthazar, Quentin, Jebediah, and Lancer, and you want to have 3 fly your sleigh. You always

have your reindeer fly in a single-file line. How many different ways can you arrange your reindeer?
Mathematics
2 answers:
Allisa [31]3 years ago
8 0

Answer:

360

Step-by-step explanation:

velikii [3]3 years ago
5 0

Answer:

120 ways.

Step-by-step explanation:

We have been given that you have 6 reindeer, Prancer, Rudy, Balthazar, Quentin, Jebediah, and Lancer, and you want to have 3 fly your sleigh. You always have your reindeer fly in a single-file line.

We will use permutation formula to solve our given problem as:

_{r}^{n}\textrm{C}=\frac{n!}{(n-r)!}

_{3}^{6}\textrm{C}=\frac{6!}{(6-3)!}

_{3}^{6}\textrm{C}=\frac{6\cdot 5\cdot 4\cdot 3!}{3!}

_{3}^{6}\textrm{C}=6\cdot 5\cdot 4

_{3}^{6}\textrm{C}=120

Therefore, you can arrange your reindeer in 120 different ways.

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Part A: 100x=235

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Mrs. Jones is building wooden picture frames. She has 1 5/8 of wood and needs 3/4 for each frame. How many frames can she make w
jonny [76]

Answer:

2\frac{1}{6}\ frames

Step-by-step explanation:

we know that

To find out how many frames she can  make with the exiting material, divide 1 5/8 by 3/4

so

1\frac{5}{8}:\frac{3}{4}

Convert mixed number to an improper fraction

1\frac{5}{8}=\frac{1*8+5}{8}=\frac{13}{8}

substitute

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6 0
3 years ago
Alice and Bob play a game involving a circle whose circumference is divided by 12 equally-spaced points. The points are numbered
inn [45]

Answer:

Step-by-step explanation:

2005 AMC 8 Problems/Problem 20

Problem

Alice and Bob play a game involving a circle whose circumference is divided by 12 equally-spaced points. The points are numbered clockwise, from 1 to 12. Both start on point 12. Alice moves clockwise and Bob, counterclockwise. In a turn of the game, Alice moves 5 points clockwise and Bob moves 9 points counterclockwise. The game ends when they stop on the same point. How many turns will this take?

$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 24$

Solution

Alice moves $5k$ steps and Bob moves $9k$ steps, where $k$ is the turn they are on. Alice and Bob coincide when the number of steps they move collectively, $14k$, is a multiple of $12$. Since this number must be a multiple of $12$, as stated in the previous sentence, $14$ has a factor $2$, $k$ must have a factor of $6$. The smallest number of turns that is a multiple of $6$ is $\boxed{\textbf{(A)}\ 6}$.

See Also

2005 AMC 8 (Problems • Answer Key • Resources)

Preceded by

Problem 19 Followed by

Problem 21

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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